Permutations and Combinations

The given digits are 1, 2, 1, 2, 0, 2 (Total 6 digits: two 1's, three 2's, one 0).
Total arrangements without leading zero restriction = $\frac{6!}{2! 3!} = 60$.
Arrangements starting with 0 = $\frac{5!}{2! 3!} = 10$.
Valid 6-digit numbers = $60 - 10 = 50$.
A number is divisible by 5 if its last digit is 0.
Number of valid 6-digit numbers ending with 0 = $\frac{5!}{2! 3!} = 10$.
Number of 6-digit numbers NOT divisible by 5 = $50 - 10 = 40$.
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(i) For 1-digit numbers (1 to 9): None have at least 2 identical digits.
(ii) For 2-digit numbers (11 to 99): Numbers with both digits same are 11, 22, ..., 99 (9 numbers).
(iii) For 3-digit numbers (100 to 999): Total 3-digit numbers = 900.
3-digit numbers with all digits distinct = $9 \times 9 \times 8 = 648$.
3-digit numbers with at least two digits same = $900 - 648 = 252$.
Total registration numbers with at least two identical digits = $9 + 252 = 261$.
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Case 1: First digit is 9.
The remaining two 9's can be placed in 2 of the remaining 6 positions in $^6C_2$ ways.
The other 4 positions can be filled with any digit except 9 (9 choices each) in $9^4$ ways.
Ways = $^6C_2 \times 9^4 = 15 \times 6561 = 98415$.
Case 2: First digit is NOT 9.
The first digit can be chosen from 1 to 8 (8 choices).
The three 9's can be placed in 3 of the remaining 6 positions in $^6C_3$ ways.
The remaining 3 positions can be filled with any of 9 digits in $9^3$ ways.
Ways = $8 \times ^6C_3 \times 9^3 = 8 \times 20 \times 729 = 116640$.
Total 7-digit numbers = $98415 + 116640 = 215055$.
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Total unrestricted arrangements of 18 balls (5 black, 7 red, 6 green) = $\frac{18!}{5! 7! 6!}$.
The opposite condition is where ALL balls of the same color are together (consecutive).
Treating the 5 black balls as 1 block, 7 red balls as 1 block, and 6 green balls as 1 block gives 3 blocks.
The number of ways to arrange these 3 blocks = $3! = 6$.
(Since balls of the same color are identical, internal arrangements within each block is 1 way).
Therefore, required arrangements = $\frac{18!}{5! 7! 6!} - 6$.
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The letters of CRICKET in alphabetical order are C, C, E, I, K, R, T.
1. Words starting with C:
* Second letter C, E, I, or K (4 choices): remaining 5 positions filled in $5! = 120$ ways.
$4 \times 5! = 480$.
2. Words starting with CR:
* Third letter C or E (2 choices): remaining 4 positions filled in $4! = 24$ ways.
$2 \times 4! = 48$.
3. Words starting with CRI:
* Fourth letter C:
* Fifth letter E: remaining 2 letters (K, T) in $2! = 2$ ways.
4. Next word in order is CRICKET itself (+1).
Total rank = $480 + 48 + 2 + 1 = 531$.

A wins the series if A reaches 4 points first. Note that in all cases, the last match played must be won by A.
Case 1: A wins all the games in a row (4 games played).
A wins the 4th game, and among the first 3 games, A wins 3 and B wins 0.
Number of ways = 1.
Case 2: A wins in 5 games.
A wins the 5th game. Of the remaining 4 games played, A wins 3 and B wins 1.
Number of ways = $\frac{4!}{3! 1!} = 4$.
Case 3: A wins in 6 games.
A wins the 6th game. Of the remaining 5 games played, A wins 3 and B wins 2.
Number of ways = $\frac{5!}{3! 2!} = 10$.
Case 4: A wins in 7 games.
A wins the 7th game. Of the remaining 6 games played, A wins 3 and B wins 3.
Number of ways = $\frac{6!}{3! 3!} = 20$.
Thus, the total number of ways in which the series can be won by A is $1 + 4 + 10 + 20 = 35$.

The word CONSTITUTION has 12 letters in total.
The occurrences of each letter are: C (1), O (2), N (2), S (1), T (3), I (2), U (1).
The total number of permutations is given by:
$\frac{12!}{1! 2! 2! 1! 3! 2! 1!} = \frac{479001600}{2 \times 2 \times 6 \times 2} = \frac{479001600}{48} = 9979200$.
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(a) ZEBRA: 5 distinct letters $\implies 5! = 120$.
(b) BAZAR: 5 letters with 2 A's $\implies \frac{5!}{2!} = 60$.
(c) SILICIC: 7 letters with 3 I's, 2 C's $\implies \frac{7!}{3! 2!} = \frac{5040}{12} = 420$.
(d) ABRACADABRA: 11 letters with 5 A's, 2 B's, 2 R's, 1 C, 1 D $\implies \frac{11!}{5! 2! 2!} = \frac{39916800}{120 \times 4} = 83160$.
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Total bulbs $n = 10$.
Bulbs of same type are identical: 5 roses, 3 daffodils, 2 sunflowers.
The number of distinct arrangements in a line is:
$\frac{10!}{5! 3! 2!} = \frac{3628800}{120 \times 6 \times 2} = \frac{3628800}{1440} = 2520$.
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Total 6 digits: one 2, two 3's, three 4's.
Total numbers formed $= \frac{6!}{1! 2! 3!} = 60$.
To find the contribution of digits at any place value:
* Number of times digit 2 is at a given place $= \frac{5!}{2! 3!} = 10$.
* Number of times digit 3 is at a given place $= \frac{5!}{1! 1! 3!} = 20$.
* Number of times digit 4 is at a given place $= \frac{5!}{1! 2! 2!} = 30$.
Sum of digits at any place $= (2 \times 10) + (3 \times 20) + (4 \times 30) = 20 + 60 + 120 = 200$.
Total sum $= 200 \times 111111 = 2222220$.
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Total items $= 4n$, consisting of $n$ identical items and $3n$ distinct items.
We need to select and arrange $2n$ items such that all $n$ identical items are included.
This requires picking $2n - n = n$ distinct items from the $3n$ available distinct items.
Number of ways to choose $n$ distinct items $= ^{3n}C_n = \frac{(3n)!}{n! (2n)!}$.
Once chosen, we have $n$ identical items and $n$ distinct items (total $2n$ items).
Number of ways to arrange these $2n$ items $= \frac{(2n)!}{n!}$.
Total permutations $= \frac{(3n)!}{n! (2n)!} \times \frac{(2n)!}{n!} = \frac{(3n)!}{(n!)^2}$.
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Each die yields an outcome from $\{1, 2, 3, 4, 5, 6\}$.
The values 2, 3, 4, 5, 6 occur twice each, using up $5 \times 2 = 10$ dice.
The remaining $12 - 10 = 2$ dice must both show the value 1.
So the 12 outcomes consist of two 1's, two 2's, two 3's, two 4's, two 5's, and two 6's.
The total number of outcomes is:
$\frac{12!}{2! 2! 2! 2! 2! 2!} = \frac{12!}{(2!)^6}$.
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Out of 100 tosses, we need exactly 52 outcomes to be heads (successes) and the remaining 48 to be tails (failures).
The number of distinct sequences of 100 tosses containing 52 heads is given by choosing 52 positions out of 100:
$^{100}C_{52} = \frac{100!}{52! 48!}$.
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Total digits $= 7$ (one 1, three 2's, two 3's, one 5).
Total 7-digit numbers $= \frac{7!}{1! 3! 2! 1!} = \frac{5040}{12} = 420$.
For odd numbers, the last digit must be odd ($1, 3, 5$).
* Ending in 1: remaining digits (three 2's, two 3's, one 5) $\implies \frac{6!}{3! 2! 1!} = 60$.
* Ending in 3: remaining digits (one 1, three 2's, one 3, one 5) $\implies \frac{6!}{1! 3! 1! 1!} = 120$.
* Ending in 5: remaining digits (one 1, three 2's, two 3's) $\implies \frac{6!}{1! 3! 2!} = 60$.
Total odd numbers $= 60 + 120 + 60 = 240$.
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For a family of $k$ children, each child can be either M or F (2 choices per child).
Number of family types with $k$ children $= 2^k$.
We need total types for families with $k = 1, 2, 3, 4, 5, 6, 7$ children:
Total $= 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7$
Total $= 2 + 4 + 8 + 16 + 32 + 64 + 128 = 254$.
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Letters of SURITI in alphabetical order: I, I, R, S, T, U.
1. Words starting with I:
Remaining letters: I, R, S, T, U (all 5 distinct) $\implies 5! = 120$.
2. Words starting with R:
Remaining letters: I, I, S, T, U (5 letters with two I's) $\implies \frac{5!}{2!} = 60$.
3. Words starting with S:
* Words starting with SI: remaining letters I, R, T, U $\implies 4! = 24$.
* Words starting with SR: remaining letters I, I, T, U $\implies \frac{4!}{2!} = 12$.
* Words starting with ST: remaining letters I, I, R, U $\implies \frac{4!}{2!} = 12$.
* Words starting with SU:
* Words starting with SUI: remaining letters I, R, T $\implies 3! = 6$.
* Words starting with SUR:
* Words starting with SURII: remaining letter T $\implies 1! = 1$ (SURIIT).
* Next word is SURITI $\implies 1$.
Total rank $= 120 + 60 + 24 + 12 + 12 + 6 + 1 + 1 = 236$.

The given set is $\{a, b, c, d, e\}$, which consists of 5 letters (vowels: $a, e$; consonants: $b, c, d$).
Case I: Repetitions allowed.
Total number of 3-letter words = $5 \times 5 \times 5 = 125$.
Number of words without any vowel (using only $b, c, d$) = $3 \times 3 \times 3 = 27$.
Number of words with at least one vowel = $125 - 27 = 98$.
Case II: Repetitions not allowed.
Total number of 3-letter words = $^5P_3 = 5 \times 4 \times 3 = 60$.
Number of words without any vowel = $^3P_3 = 3 \times 2 \times 1 = 6$.
Number of words with at least one vowel = $60 - 6 = 54$.

There are 6 letters in the word PENCIL (2 vowels: E, I; 4 consonants: P, N, C, L).
The 6 places are 1st, 2nd, 3rd, 4th, 5th, and 6th, among which the even places are 2nd, 4th, and 6th (3 places).
The 2 vowels can be arranged in 3 available even places in $^3P_2 = 6$ ways.
The remaining 4 consonants can be arranged in the remaining 4 places in $4! = 24$ ways.
Total required number of words = $6 \times 24 = 144$.
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The word ENGLISH consists of 7 distinct letters.
(i) Fixing E at the first position leaves 6 remaining letters to be arranged in 6 spaces, which can be done in $6! = 720$ ways.
(ii) Fixing H at the last position leaves 6 remaining letters to be arranged in 6 spaces, which can be done in $6! = 720$ ways.
(iii) Fixing E at the first position and H at the last position leaves 5 remaining letters to be arranged in the 5 middle spaces, which can be done in $5! = 120$ ways.
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The middle digit (tens place) can be chosen from the perfect squares in $\{1, 2, \dots, 9\}$, which are 1, 4, and 9 (3 choices).
The hundreds place can be filled by any of the 9 digits ($1$ to $9$) in 9 ways.
The units place can be filled by any of the 9 digits ($1$ to $9$) in 9 ways.
Total such numbers = $9 \times 3 \times 9 = 243$.
Sum contributed by hundreds place:
Each digit $1, 2, \dots, 9$ appears $3 \times 9 = 27$ times in the hundreds place.
Sum = $27 \times (1 + 2 + \dots + 9) \times 100 = 27 \times 45 \times 100 = 121500$.
Sum contributed by tens place:
Each perfect square digit ($1, 4, 9$) appears $9 \times 9 = 81$ times in the tens place.
Sum = $81 \times (1 + 4 + 9) \times 10 = 81 \times 14 \times 10 = 11340$.
Sum contributed by units place:
Each digit $1, 2, \dots, 9$ appears $9 \times 3 = 27$ times in the units place.
Sum = $27 \times (1 + 2 + \dots + 9) \times 1 = 27 \times 45 = 1215$.
Total sum = $121500 + 11340 + 1215 = 134055$.
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There are $2 + 3 = 5$ distinct book titles in total.
Since all copies of the same book must stay together, treat all copies of each book as 1 single block/unit.
This gives 5 units to arrange in a row, which can be done in $5! = 120$ ways.
Since copies of the same book are identical, their internal arrangements within each unit can be done in only 1 way.
Therefore, total required arrangements = $5! = 120$.
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Regarding all persons of the same nationality as one single block/unit, there are 3 units (Indians, Americans, Englishmen).
These 3 units can be arranged among themselves in $3!$ ways.
Within their respective groups:
* 8 Indians can be arranged among themselves in $8!$ ways.
* 4 Americans can be arranged among themselves in $4!$ ways.
* 4 Englishmen can be arranged among themselves in $4!$ ways.
Using the multiplication principle, total number of arrangements = $3! \times 8! \times 4! \times 4!$.

Total number of letters in UNIVERSITY = 10.
The vowels are U, I, E, I (total 4 vowels, where 'I' occurs twice).
Regarding the 4 vowels as a single block, we have 6 remaining letters + 1 vowel block = 7 blocks.
These 7 blocks can be arranged in $7!$ ways.
Within the vowel block, the 4 vowels (with 2 I's) can be arranged among themselves in $\frac{4!}{2!}$ ways.
Total required arrangements = $7! \times \frac{4!}{2!} = 5040 \times 12 = 60480$.
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Let the two specified speakers be A and B.
(i) Five persons can be arranged without restriction in $5! = 120$ ways. In half of these arrangements A speaks before B, and in the other half B speaks before A. Thus, required ways = $\frac{1}{2} \times 5! = 60$.
(ii) Regarding AB (in that exact order) as a single unit, we arrange this unit along with the remaining 3 persons (total 4 units). The number of arrangements is $4! = 24$.
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First, arrange the 10 IIT students in $10!$ ways.
To have exactly 3 IIT students between the 2 PET students, consider the group of 3 consecutive IIT students.
Out of 10 positions of IIT students, a block of 3 consecutive IIT students can be selected in 8 ways.
The 2 PET students can be placed at the two ends of this block in $2! = 2$ ways.
Thus, total number of arrangements = $10! \times 8 \times 2! = 16 \times 10!$.
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The total number of ways to seat 10 students without any restriction is $10!$.
If the 3 girls sit together, treat them as a single block. This leaves 7 boys + 1 girl block = 8 units to arrange, which can be done in $8!$ ways.
The 3 girls can be arranged among themselves in $3!$ ways.
Ways where all 3 girls sit together = $8! \times 3!$.
Hence, the number of ways such that all three girls DO NOT sit together = $10! - 8! \times 3!$.
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Total permutations of $n$ elements = $n!$.
To find the number of permutations where $a$ and $b$ ARE adjacent, treat $(a, b)$ as a single block.
This leaves $n - 1$ elements to arrange in $(n - 1)!$ ways, and $a$ and $b$ can be arranged within the block in $2! = 2$ ways.
So, adjacent arrangements = $2 \times (n - 1)!$.
Permutations where $a$ and $b$ are NOT adjacent = $n! - 2(n - 1)! = n(n - 1)! - 2(n - 1)! = (n - 1)! (n - 2)$.

First, arrange the 6 non-mathematics candidates in a row. They can be arranged in $6! = 720$ ways.
This creates 7 available gaps (5 between candidates and 2 at the ends) for placing the mathematics candidates.
The 4 mathematics candidates can be placed into these 7 gaps in $^7P_4 = 7 \times 6 \times 5 \times 4 = 840$ ways.
Total number of seating arrangements = $6! \times ^7P_4 = 720 \times 840 = 604800$.
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There are 10 multiples of 3 (namely $3, 6, 9, \dots, 30$) and 20 non-multiples of 3.
First, arrange the 20 non-multiples of 3 in a row, which can be done in $20!$ ways.
This creates 21 gaps (including the two ends).
We need to select 10 of these 21 gaps for the multiples of 3, which can be done in $^{21}C_{10}$ ways.
Once selected, the 10 multiples of 3 can be arranged among themselves in $10!$ ways.
Total ways = $20! \times ^{21}C_{10} \times 10! = 20! \times \frac{21!}{10! 11!} \times 10! = \frac{20! \times 21!}{11!}$.
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(i) Treat the two B's as a single object (BB). The remaining non-A objects are (BB), C, and D (3 items).
These 3 items can be arranged in $3! = 6$ ways, creating 4 available gaps.
The three A's can be placed in any 3 of these 4 gaps in $^4C_3 = 4$ ways.
Total ways for (i) = $3! \times ^4C_3 = 6 \times 4 = 24$.
(ii) First, find the total arrangements where no two A's are together regardless of B's.
The non-A objects are B, B, C, D (4 items with two identical B's).
Arrangements of B, B, C, D = $\frac{4!}{2!} = 12$ ways, creating 5 gaps.
Placing three A's in these 5 gaps = $^5C_3 = 10$ ways.
Total arrangements with no two A's together = $12 \times 10 = 120$.
Subtracting the cases from part (i) where the two B's ARE together:
Total ways for (ii) = $120 - 24 = 96$.
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The word INDEPENDENCE has 12 letters: 3 N's, 4 E's, 2 D's, 1 I, 1 P, 1 C.
Total unrestricted arrangements = $\frac{12!}{3! 4! 2!} = 1663200$.
(i) Fixing P at the first position leaves 11 letters (3 N's, 4 E's, 2 D's, 1 I, 1 C):
Arrangements starting with P = $\frac{11!}{3! 4! 2!} = 138600$.
(ii) Vowels are E, E, E, E, I (5 vowels). Group all 5 vowels into 1 block.
This block together with the 7 consonants (3 N's, 2 D's, 1 P, 1 C) forms 8 objects.
Arranging these 8 objects (with 3 N's and 2 D's) = $\frac{8!}{3! 2!}$ ways.
Arranging the 5 vowels inside their block (with 4 E's) = $\frac{5!}{4!}$ ways.
Total ways with vowels together = $\frac{8!}{3! 2!} \times \frac{5!}{4!} = 3360 \times 5 = 16800$.
(iii) Total arrangements where vowels never occur together:
Unrestricted total - Vowels together = $1663200 - 16800 = 1646400$.

First, calculate the total number of persons, which is $5 + 3 = 8$.
The total number of unrestricted arrangements of 8 persons in a row is $8!$.
When all 3 girls sit together, consider them as a single unit. This gives $5 + 1 = 6$ units to arrange.
These 6 units can be arranged in $6!$ ways.
The 3 girls within their unit can be arranged among themselves in $3!$ ways.
Therefore, the number of arrangements where all three girls sit together is $6! \times 3!$.
Subtracting this from the total number of arrangements gives the required number of ways:
$\text{Required ways} = 8! - 6! \times 3!$
$\text{Required ways} = 6! \times (8 \times 7 - 6)$
$\text{Required ways} = 720 \times (56 - 6)$
$\text{Required ways} = 720 \times 50 = 36000$

Case I: When a boy occupies the first place, the seating pattern is B G B G B G B G.
The 4 boys can be seated in 4 places in $4!$ ways.
The 4 girls can be seated in 4 places in $4!$ ways.
Number of arrangements for Case I = $4! \times 4! = 576$.
Case II: When a girl occupies the first place, the seating pattern is G B G B G B G B.
The 4 girls can be seated in 4 places in $4!$ ways.
The 4 boys can be seated in 4 places in $4!$ ways.
Number of arrangements for Case II = $4! \times 4! = 576$.
Total number of arrangements = $4! \times 4! + 4! \times 4! = 2 \times 4! \times 4! = 2 \times 24 \times 24 = 1152$.

Since there are 4 boys and 3 girls, the only possible arrangement where they sit alternately is B G B G B G B.
The 4 boys can be seated in the 4 odd places (1st, 3rd, 5th, and 7th) in $4!$ ways.
The 3 girls can be seated in the 3 even places (2nd, 4th, and 6th) in $3!$ ways.
Total number of required ways = $4! \times 3! = 24 \times 6 = 144$.

The arrangement must follow the pattern B G B G B G B.
This single configuration allows the 4 boys to be arranged in $4!$ ways and the 3 girls to be arranged in $3!$ ways.
Hence, the number of ways = $4! \times 3! = 24 \times 6 = 144$.

Case I: A boy occupies the first place (B G B G B G B G).
The 4 boys and 4 girls have 4 places each.
Number of ways = $4! \times 4! = 576$.
Case II: A girl occupies the first place (G B G B G B G B).
Number of ways = $4! \times 4! = 576$.
Hence, total number of ways = $2 \times 4! \times 4! = 2 \times 576 = 1152$.
Total number of three-letter words = $31 + 32 = 63$.

The letters in the word 'Keppelin' are K, e, p, p, e, l, i, n (total 8 letters: K, e, e, p, p, l, i, n).
The available vowels are 'e' and 'i'.
Case I: When 'e' is in the middle position.
Excluding one 'e', the remaining available letters are K, e, p, p, l, i, n (6 distinct types: K, e, p, l, i, n, with 'p' occurring twice).
Number of ways with 2 different letters at the ends:
Choose 2 distinct letters from 6 available types in $^6C_2$ ways and arrange them in $2!$ ways.
$^6C_2 \times 2! = 15 \times 2 = 30$.
Number of ways with 2 like letters at the ends (p e p):
Only 1 way (using the two 'p's).
Total words with 'e' in the middle = $30 + 1 = 31$.
Case II: When 'i' is in the middle position.
Excluding 'i', the remaining available letters are K, e, e, p, p, l, n (5 distinct types: K, e, p, l, n, with 'e' occurring twice and 'p' occurring twice).
Number of ways with 2 different letters at the ends:
Choose 2 distinct letters from 5 available types in $^5C_2$ ways and arrange them in $2!$ ways.
$^5C_2 \times 2! = 10 \times 2 = 20$.
Number of ways with 2 like letters at the ends (p i p, e i e):
There are 2 ways (using either the two 'p's or the two 'e's).
Total words with 'i' in the middle = $20 + 2 = 22$.
Total number of three-letter words = $31 + 22 = 53$.

There are 7 positions in total, with 4 odd places (1st, 3rd, 5th, 7th) and 3 even places (2nd, 4th, 6th).
The given digits are 1, 2, 3, 4, 3, 2, 1.
Odd digits: 1, 1, 3, 3 (4 digits: two 1s and two 3s).
Even digits: 2, 2, 4 (3 digits: two 2s and one 4).
The 4 odd digits can be arranged in the 4 odd places in $\frac{4!}{2! \times 2!}$ ways:
$\frac{4!}{2! \times 2!} = \frac{24}{4} = 6$ ways.
The 3 even digits can be arranged in the 3 even places in $\frac{3!}{2!}$ ways:
$\frac{3!}{2!} = \frac{6}{2} = 3$ ways.
Total required numbers = $6 \times 3 = 18$.

Total men = 11.
Steering specialists = 3.
Rowers = $11 - 3 = 8$.
Step 1: Select 1 person out of 3 to steer the boat in $^3C_1 = 3$ ways.
Step 2: Two particular men must row on the bow side. They can be placed in 2 out of 4 places on the bow side in $^4P_2 = 12$ ways.
Step 3: From the remaining 6 men, select 2 men for the remaining 2 places on the bow side in $^6C_2 = 15$ ways.
Step 4: The remaining 4 men occupy the 4 places on the stroke side, which can be arranged in $4! = 24$ ways.
Total required arrangements = $^3C_1 \times ^4P_2 \times ^6C_2 \times 4! = 3 \times 12 \times 15 \times 24 = 25920$.

The letters of the word SUCCESS are S, U, C, C, E, S, S (3 S's, 2 C's, 1 U, 1 E; total 7 letters).
Part (a): Two C are together but no two S are together.
Treat CC as a single object.
First, arrange the remaining non-S letters, which are U, CC, E (3 distinct objects).
Number of ways to arrange U, CC, E = $3! = 6$.
These 3 objects create 4 gap positions: $\text{\_ U \_ CC \_ E \_}$.
Place the 3 identical S's into 3 of these 4 gaps in $^4C_3 = 4$ ways.
Total ways for (a) = $3! \times ^4C_3 = 6 \times 4 = 24$.
Part (b): No two C and no two S are together.
First find the total number of arrangements where no two S are together.
Arrange U, C, C, E in $\frac{4!}{2!} = 12$ ways.
This creates 5 gap positions for the 3 S's: $\frac{4!}{2!} \times ^5C_3 = 12 \times 10 = 120$ ways.
From these 120 arrangements, subtract those where the two C are together (which is 24 from part a):
Total ways for (b) = $120 - 24 = 96$.

Total unrestricted arrangements of 6 books = $6! = 720$.
Part (a): Two particular books must be side by side.
Treat the 2 particular books as a single unit.
Now, we have $5 - 1 + 1 = 5$ units to arrange, which can be done in $5!$ ways.
The 2 particular books can be arranged among themselves in $2!$ ways.
Number of ways = $5! \times 2! = 120 \times 2 = 240$.
Part (b): The two books must not be side by side.
Subtract the number of ways where they are side by side from the total arrangements:
Number of ways = $720 - 240 = 480$.

There are 4 distinct topics: Statistics, Algebra, Calculus, and Astronomy.
Step 1: The 4 topic groups can be arranged among themselves in $4!$ ways.
Step 2: Internal arrangement within each topic:
* 2 identical copies of Statistics book: $\frac{2!}{2!} = 1$ way.
* 2 identical copies of Algebra book: $\frac{2!}{2!} = 1$ way.
* 3 identical copies of Calculus book: $\frac{3!}{3!} = 1$ way.
* 1 copy of Astronomy book: $1! = 1$ way.
Total number of arrangements = $4! \times 1 \times 1 \times 1 \times 1 = 24$.

Total unrestricted arrangements of 30 volumes = $30!$.
Part (a): Volumes 1 and 2 are side by side.
Treat volumes 1 and 2 as a single block.
Now there are 29 units to arrange, which can be done in $29!$ ways.
Volumes 1 and 2 can be arranged within the block in $2! = 2$ ways.
Number of ways = $2 \times 29!$.
Part (b): Volumes 3 and 4 are not side by side.
Number of ways volumes 3 and 4 are side by side = $2 \times 29!$.
Number of ways where they are not side by side = $30! - (2 \times 29!)$.
Factoring out $29!$:
$30 \times 29! - 2 \times 29! = (30 - 2) \times 29! = 28 \times 29!$.

The given set of digits is {0, 1, 2, 3, 4, 5, 6, 7} (total 8 digits).
Total four-digit numbers formed from these 8 digits (with repetition allowed):
* Thousands place: 7 choices (1 through 7, as 0 cannot be the leading digit).
* Hundreds place: 8 choices.
* Tens place: 8 choices.
* Units place: 8 choices.
Total = $7 \times 8 \times 8 \times 8 = 3584$.
Four-digit numbers formed without using the digit 1 (digits available: {0, 2, 3, 4, 5, 6, 7}, total 7 digits):
* Thousands place: 6 choices (2 through 7).
* Hundreds place: 7 choices.
* Tens place: 7 choices.
* Units place: 7 choices.
Total without digit 1 = $6 \times 7 \times 7 \times 7 = 2058$.
Number of four-digit numbers containing at least one digit 1 = $3584 - 2058 = 1526$.
Alternatively, if repetition of digits is not allowed:
Case 1: Digit 1 is in the thousand's place: $1 \times 7 \times 6 \times 5 = 210$.
Case 2: Digit 1 is in one of the other 3 places (hundreds, tens, or units):
* Choose position for 1 in 3 ways.
* Thousands place: 6 choices (excluding 0 and 1).
* Remaining 2 places: $6 \times 5 = 30$ choices.
Total for Case 2 = $3 \times 6 \times 30 = 540$.
Total without repetition = $210 + 540 = 750$.
For standard four-digit numbers allowing repetition, the total containing digit 1 is $3584 - 2058 = 1526$, or using exact counts with exactly one digit 1:
Position of 1 is thousands place: $1 \times 7 \times 7 \times 7 = 343$.
Position of 1 is hundreds, tens, or units place: $3 \times (6 \times 1 \times 7 \times 7) = 3 \times 294 = 882$.
Total with exactly one digit 1 = $343 + 882 = 1225$.
If repetition is not allowed and digits are distinct:
Total = $210 + 3 \times (6 \times 6 \times 5) = 210 + 540 = 750$.
Considering the standard textbook interpretation for exactly one digit 1 with distinct digits:
Total = $750$.

The word PERMUTATIONS has 12 letters: P, E, R, M, U, T, A, T, I, O, N, S.
Letter frequencies: T occurs 2 times, all other letters occur once.
(i) Words start with P and end with S:
Fix P at 1st position and S at 12th position.
The remaining 10 letters (E, R, M, U, T, A, T, I, O, N) contain two T's.
Number of ways = $\frac{10!}{2!} = \frac{3628800}{2} = 1814400$.
(ii) Vowels are all together:
Vowels in PERMUTATIONS: E, U, A, I, O (5 vowels, all distinct).
Consonants: P, R, M, T, T, N, S (7 consonants, with two T's).
Treat the 5 vowels as 1 single block.
Total units to arrange = 7 consonants + 1 block = 8 units.
Arrangement of 8 units with two T's = $\frac{8!}{2!} = \frac{40320}{2} = 20160$.
Internal arrangement of 5 distinct vowels = $5! = 120$.
Number of ways = $20160 \times 120 = 2419200$.
(iii) Always 4 letters between P and S:
The positions for P and S can be (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12).
There are 7 such pairs of positions.
Since P and S can interchange places, there are $7 \times 2 = 14$ ways to place P and S.
The remaining 10 positions are filled by the other 10 letters (including two T's) in $\frac{10!}{2!}$ ways.
Number of ways = $14 \times \frac{10!}{2!} = 14 \times 1814400 = 25401600$.

Total number of lessons = 5.
Total unrestricted arrangements = $5! = 120$.
To find arrangements where algebra and geometry immediately follow each other:
Treat algebra and geometry as a single unit.
This leaves 4 units to arrange, which can be done in $4! = 24$ ways.
Algebra and geometry can be arranged among themselves within the unit in $2! = 2$ ways.
Number of ways where they follow each other = $24 \times 2 = 48$.
Subtracting this from total arrangements gives the required ways:
$120 - 48 = 72$.

The given digits are 1, 2, 3, 4, 5.
Part (i): Digits can be repeated.
The last two digits must form a two-digit number divisible by 4.
Valid pairs for last two digits: 12, 24, 32, 44, 52 (5 choices).
The first three digits can each be filled in 5 ways.
Number of ways = $5 \times 5 \times 5 \times 5 = 625$.
Part (ii): Digits cannot be repeated.
Valid pairs of distinct digits for the last two positions: 12, 24, 32, 52 (4 choices).
For each choice of the last two digits, the remaining 3 positions can be filled using the remaining 3 digits in $3! = 6$ ways.
Number of ways = $4 \times 3! = 4 \times 6 = 24$.

The given digits are {0, 1, 2, 3, 4, 5}.
Sum of all 6 digits = $0 + 1 + 2 + 3 + 4 + 5 = 15$.
To choose 5 digits whose sum is divisible by 3, we must exclude one digit such that the remaining sum is a multiple of 3.
The excluded digit can be 0 or 3.
Case 1: Exclude 0 (Chosen digits: {1, 2, 3, 4, 5}).
Sum = 15 (divisible by 3).
To be even, the last digit must be 2 or 4 (2 choices).
Remaining 4 places can be filled in $4! = 24$ ways.
Number of ways for Case 1 = $2 \times 24 = 48$.
Case 2: Exclude 3 (Chosen digits: {0, 1, 2, 4, 5}).
Sum = 12 (divisible by 3).
Subcase 2a: Last digit is 0.
First 4 places filled by {1, 2, 4, 5} in $4! = 24$ ways.
Subcase 2b: Last digit is 2 or 4 (2 choices).
First digit cannot be 0 (3 choices left), and remaining 3 middle places filled in $3! = 6$ ways.
Ways for Subcase 2b = $2 \times 3 \times 6 = 36$.
Number of ways for Case 2 = $24 + 36 = 60$.
Total 5-digit numbers divisible by 6 = $48 + 60 = 108$ (Wait, re-evaluating remaining cases):
If we exclude 0, remaining digits {1,2,3,4,5} give 48 ways.
If we exclude 3, remaining digits {0,1,2,4,5} give 60 ways.
Sum = $48 + 60 = 108$.
Let's check if excluding 0 gives 120 / 3?
Actually:
Case 1: {1, 2, 3, 4, 5} -> Last digit 2 or 4: $2 \times 4! = 48$.
Case 2: {0, 1, 2, 4, 5} ->
If end is 0: $4! = 24$.
If end is 2: $3 \times 3! = 18$.
If end is 4: $3 \times 3! = 18$.
Total for Case 2 = $24 + 18 + 18 = 60$.
Total overall = $48 + 60 = 108$.
Options recalculation for 108 vs standard permutations:
Total numbers = 192 (if combined with other problem formats). Here, total is 192 if including 0, 1, 2, 3, 4, 5 across all configurations.

Available digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (10 digits).
In telephone numbers, zero can occupy the first place.
The 1st digit can be chosen in 10 ways.
The 2nd digit can be chosen in 9 ways.
The 3rd digit can be chosen in 8 ways.
The 4th digit can be chosen in 7 ways.
The 5th digit can be chosen in 6 ways.
Total number of telephone numbers = $^{10}P_5 = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.

Available digits = {0, 1, 2, 3, 4, 5, 6, 7} (8 digits).
For a number to be divisible by 25, the last two digits must be formed by 25, 50, or 75 (assuming distinct digits).
Case 1: Ending with 50 (last two digits are 5 and 0).
The remaining 3 positions (1st, 2nd, 3rd places) must be filled from the remaining 6 digits {1, 2, 3, 4, 6, 7}.
Since 0 is already used at the end, any remaining digit can be in the 1st place.
Number of ways = $^6P_3 = 6 \times 5 \times 4 = 120$.
Case 2: Ending with 25 or 75 (2 possibilities).
The remaining 6 available digits include 0.
The 1st digit cannot be 0, so there are 5 choices for the 1st place.
The 2nd place can be filled in 5 ways (including 0).
The 3rd place can be filled in 4 ways.
Number of ways for each ending = $5 \times 5 \times 4 = 100$.
Total for Case 2 = $2 \times 100 = 200$.
Total 5-digit numbers without repetition = $120 + 200 = 320$.
If repetition of digits is allowed:
Ending pairs: 00, 25, 50, 75 (4 choices).
For 00, 25, 50, 75:
1st place: 7 choices (excluding 0).
2nd place: 8 choices.
3rd place: 8 choices.
Total = $4 \times (7 \times 8 \times 8) = 4 \times 448 = 1792$.