Permutations and Combinations

Total unrestricted arrangements of the 4 letters $a, b, c, d$ is $4! = 24$.
Let $X$ be the set of arrangements where $a, b$ are together.
Treating $(ab)$ as a single unit, there are 3 units to arrange in $3! = 6$ ways.
Since $a$ and $b$ can swap places in $2! = 2$ ways:
$\vert{}X\vert{} = 3! \times 2! = 6 \times 2 = 12$.
Let $Y$ be the set of arrangements where $c, d$ are together.
Similarly, $\vert{}Y\vert{} = 3! \times 2! = 12$.
Now find $\vert{}X \cap Y\vert{}$, the arrangements where both $a, b$ are together and $c, d$ are together.
Treating $(ab)$ as one unit and $(cd)$ as another unit, there are 2 units to arrange in $2! = 2$ ways.
$a, b$ can be arranged in $2!$ ways and $c, d$ in $2!$ ways:
$\vert{}X \cap Y\vert{} = 2! \times 2! \times 2! = 8$.
By Principle of Inclusion-Exclusion, the number of arrangements where at least one pair is together is:
$\vert{}X \cup Y\vert{} = \vert{}X\vert{} + \vert{}Y\vert{} - \vert{}X \cap Y\vert{} = 12 + 12 - 8 = 16$.
The number of arrangements where neither pair comes together is:
$\text{Required ways} = 24 - 16 = 8$.

An 8-oared boat has 4 positions on the bow side and 4 positions on the stroke side.
There are 8 men in total:
* 2 men must row on the bow side.
* 1 man must row on the stroke side.
* 5 men can row on either side.
Step 1: Place the 2 restricted bow rowers in 2 of the 4 bow positions in $^4P_2 = 12$ ways.
Step 2: Place the 1 restricted stroke rower in 1 of the 4 stroke positions in $^4P_1 = 4$ ways.
Step 3: Now 2 bow positions and 3 stroke positions remain open (total 5 positions).
The remaining 5 unrestricted men can be arranged in these 5 remaining positions in $5! = 120$ ways.
Total number of arrangements = $^4P_2 \times ^4P_1 \times 5! = 12 \times 4 \times 120 = 5760$.

There are 3 nationality groups:
* Group 1: 10 Indians
* Group 2: 5 Americans
* Group 3: 5 Englishmen
Step 1: Treat each nationality group as 1 single block. The 3 blocks can be arranged in $3! = 6$ ways.
Step 2: Arrange the individuals within each group:
* 10 Indians can sit among themselves in $10!$ ways.
* 5 Americans can sit among themselves in $5!$ ways.
* 5 Englishmen can sit among themselves in $5!$ ways.
Total number of arrangements = $3! \times 10! \times 5! \times 5!$.

Total papers = 6.
Total unrestricted ways to set 6 papers = $6! = 720$.
To find the number of ways where the 2 mathematics papers are successive (together):
Treat the 2 mathematics papers as a single block.
Now there are $5 - 1 + 1 = 5$ units to arrange, which can be done in $5! = 120$ ways.
The 2 mathematics papers within the block can be arranged in $2! = 2$ ways.
Number of ways where mathematics papers are successive = $5! \times 2! = 120 \times 2 = 240$.
Subtracting this from total arrangements gives:
Number of required orders = $720 - 240 = 480$.

The word 'PLANTAIN' has 8 letters:
Letters count: P (1), L (1), A (2), N (2), T (1), I (1).
First, arrange the non-A letters: P, L, N, N, T, I (6 letters, with 2 N's).
Number of ways to arrange these 6 letters = $\frac{6!}{2!} = \frac{720}{2} = 360$.
Arranging these 6 letters creates 7 gap positions (before the first, between letters, and after the last):
$\text{\_ L\_ N \_ T \_ I \_ N \_ P \_}$
To ensure the two 'A's are not together, choose 2 gap positions out of the 7 available gaps for the two identical 'A's in $^7C_2$ ways:
$^7C_2 = \frac{7 \times 6}{2} = 21$ ways.
Total required arrangements = $360 \times 21 = 3780$.

The word 'SALOON' has 6 letters consisting of:
* Consonants: S, L, N (3 distinct consonants)
* Vowels: A, O, O (3 vowels with 2 identical O's)
Since there are 3 consonants and 3 vowels, the alternating arrangement can take two forms:
Case 1: Consonant first (C V C V C V).
* Arrange 3 distinct consonants in 3 places: $3! = 6$ ways.
* Arrange 3 vowels (A, O, O) in 3 places: $\frac{3!}{2!} = 3$ ways.
* Number of ways for Case 1 = $6 \times 3 = 18$.
Case 2: Vowel first (V C V C V C).
* Arrange 3 vowels (A, O, O) in 3 places: $\frac{3!}{2!} = 3$ ways.
* Arrange 3 consonants in 3 places: $3! = 6$ ways.
* Number of ways for Case 2 = $3 \times 6 = 18$.
Total number of alternate arrangements = $18 + 18 = 36$.

The word 'BHARAT' consists of 6 letters: B, H, A, R, A, T (with 2 A's).
Part 1: Total words formed.
Total permutations = $\frac{6!}{2!} = \frac{720}{2} = 360$.
Part 2: Words that will NOT contain B and H together.
Treat (BH) as a single block.
The remaining units are (BH), A, R, A, T (5 units with 2 A's).
Arrangements with B and H together = $\frac{5!}{2!} \times 2! = \frac{120}{2} \times 2 = 120$.
Words not containing B and H together = $360 - 120 = 240$.
Part 3: Words starting with B and ending with T.
Fix B at the 1st position and T at the 6th position.
The remaining 4 middle positions are filled by H, A, R, A (4 letters with 2 A's).
Number of arrangements = $\frac{4!}{2!} = \frac{24}{2} = 12$.
Summary: Total = 360, Without BH together = 240, Starting with B and ending with T = 12.

There are 9 distinct books: 4 red (R) and 5 green (G).
(a) No restrictions:
The 9 distinct books can be arranged in $9! = 362880$ ways.
(b) Red books together and green books together:
Treat all 4 red books as 1 unit and all 5 green books as 1 unit.
The 2 units can be arranged in $2! = 2$ ways.
The 4 distinct red books can be arranged among themselves in $4! = 24$ ways.
The 5 distinct green books can be arranged among themselves in $5! = 120$ ways.
Total ways = $2! \times 4! \times 5! = 2 \times 24 \times 120 = 5760$.
(c) Red books must be together:
Treat the 4 red books as a single unit.
Total units to arrange = 5 green books + 1 red unit = 6 units.
The 6 units can be arranged in $6! = 720$ ways.
The 4 red books can be arranged among themselves in $4! = 24$ ways.
Total ways = $6! \times 4! = 720 \times 24 = 28800$.
(d) Colours must alternate:
Since there are 5 green books and 4 red books, the only valid alternating pattern is G R G R G R G R G.
The 5 green books can be arranged in the 5 odd places in $5! = 120$ ways.
The 4 red books can be arranged in the 4 even places in $4! = 24$ ways.
Total ways = $5! \times 4! = 120 \times 24 = 2880$.

The number of ways of selecting $5$ consecutive card denominations from $13$ available values in a suit is given by $n - r + 1 = 13 - 5 + 1 = 9$. Since a card of any denomination can be selected from any of the $4$ suits in $4$ ways, $5$ cards of five consecutive denominations can be selected in $4^5 = 1024$ suit combinations. Therefore, the total number of different straights is $9 \times 1024 = 9216$. The number of ways to select $5$ consecutive cards belonging to the exact same suit is $9$ per suit, giving $9 \times 4 = 36$ total flush straights. Subtracting these gives $9216 - 36 = 9180$ straights where the cards are not all from the same suit.

(i) If the $n$ objects are $a_1, a_2, a_3, \dots, a_n$, the possible consecutive selections of three objects are $(a_1, a_2, a_3), (a_2, a_3, a_4), \dots, (a_{n-2}, a_{n-1}, a_n)$. The total number of such selections is $n - 3 + 1 = n - 2$.
(ii) The total number of ways to select $3$ objects out of $n$ without any restriction is $^nC_3 = \frac{n(n - 1)(n - 2)}{6}$. Subtracting the number of consecutive selections from the total selections gives $\frac{n(n - 1)(n - 2)}{6} - (n - 2) = \frac{n - 2}{6}[n(n - 1) - 6] = \frac{(n - 2)(n^2 - n - 6)}{6}$.

(i) Let the intermediate stations be $S_1, S_2, S_3, \dots, S_n$. The number of triplets of consecutive stations $(S_1 S_2 S_3, S_2 S_3 S_4, \dots, S_{n-2} S_{n-1} S_n)$ is $n - 3 + 1 = n - 2$. (ii) The total number of consecutive pairs of stations $(S_1 S_2, S_2 S_3, \dots, S_{n-1} S_n)$ is $n - 1$. Each pair can be associated with any of the remaining $(n - 2)$ stations to form a triplet with at least two consecutive stations, giving $(n - 1)(n - 2)$ ways. However, cases where all three stations are consecutive (e.g., $S_4 S_5$ combined with $S_6$, and $S_5 S_6$ combined with $S_4$) are counted twice. Subtracting the excess count of these $n - 2$ triplets gives $(n - 1)(n - 2) - (n - 2) = (n - 2)^2$.

(i) The total number of consecutive pairs of vertices (sides of the polygon) $A_1 A_2, A_2 A_3, \dots, A_n A_1$ is $n$. Each pair can be associated with a third vertex in $n - 3$ ways to form a triangle sharing at least one side, avoiding double counting of triangles sharing two sides. Thus, the number of triangles having at least one side common with the polygon is $n(n - 3)$.
(ii) The total number of triangles formed by choosing any $3$ vertices out of $n$ without restrictions is $^nC_3 = \frac{n(n - 1)(n - 2)}{6}$. The number of triangles having no side coinciding with the polygon is obtained by subtracting the restricted triangles from the total count: $^nC_3 - n(n - 3) = \frac{n(n - 1)(n - 2) - 6n(n - 3)}{6} = \frac{n(n^2 - 3n + 2 - 6n + 18)}{6} = \frac{n(n^2 - 9n + 20)}{6} = \frac{n(n - 4)(n - 5)}{6}$.

If the stations $S_1, S_2, \dots, S_8$ are arranged in a circular order, we need to select $3$ non-adjacent stations. If a particular station, say $S_1$, is selected, then adjacent stations $S_2$ and $S_8$ cannot be selected. From the remaining $5$ stations ($S_3, S_4, S_5, S_6, S_7$), we must select $2$ non-adjacent stations. The number of ways to choose $2$ non-consecutive items from $5$ in a row is equivalent to $^4C_2 = 6$. Since any of the $8$ stations can be chosen as the starting reference point, this gives $8 \times 6 = 48$ counted combinations. However, each selection of $3$ stations is counted $3$ times (once for each of the three chosen stations). Thus, the total number of required selections is $\frac{8 \times 4}{3} = 16$. Alternatively, applying the general circular non-consecutive formula $\frac{n}{n - r} ^{n - r}C_r$ with $n = 8$ and $r = 3$ gives $\frac{8}{8 - 3} ^{8 - 3}C_3 = \frac{8}{5} \times ^5C_3 = \frac{8}{5} \times 10 = 16$.

The total number of $7$-digit numbers with distinct digits chosen from $S = \{1, 2, \dots, 9\}$ without any restrictions is given by $n_2 = ^9P_7 = 181440$. Now, consider the $7$ positions as $a, b, c, d, e, f, g$. There are $6$ pairs of consecutive places, namely $ab, bc, cd, de, ef, fg$, where the digits $5$ and $6$ can appear together. In each pair, $5$ and $6$ can be placed in $2! = 2$ ways (either as $56$ or $65$). For each such placement, the remaining $5$ places can be filled from the remaining $7$ digits in $^7P_5 = 2520$ ways. Thus, the number of $7$-digit numbers containing $56$ or $65$ consecutively is $n_1 = 2 \times 6 \times ^7P_5 = 12 \times 2520 = 30240$. Subtracting $n_1$ from $n_2$ gives the number of valid $7$-digit numbers where $5$ and $6$ do not appear consecutively: $181440 - 30240 = 151200$.