Permutations and Combinations

Number of ways to arrange 5 boys in a circle is $4!

$ \begin{aligned} & =4!\times{ }^5 C_5 \times 5! \\ & =24 \times 12 \times 10 \\ & =2880 \end{aligned} $

Remaining words excluding GENTLE are IINCEL.

$ \begin{aligned} \therefore \text { Number of ways } & =\frac{{ }^1 C_1 \times{ }^4 C_1 \times 1 \times 6!}{2!} \\ & =\frac{4 \times 720}{2}=1440 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{20!}{15!}-\frac{19!}{14!} \\ & \quad=\frac{20 \times 19!}{15 \times 14!}-\frac{19!}{14!} \\ & \quad=\frac{19!}{14!}\left(\frac{20-15}{15}\right) \\ & \quad=\frac{19 \times 18 \times 17 \times 16 \times 15 \times 5}{15} \\ & \quad=19 \times 17 \times 2 \times 9 \times 80 \\ & \quad=6!\times 646 \end{aligned} \end{aligned} $

$ \text { Arrangement }=\frac{4!}{2!} $

Selection of 3 gaps out of 5 for $3 A^{\prime} 5={ }^5 C_3$

∴ Required arrangements

$ \begin{aligned} & =\frac{4!}{2!} \cdot{ }^5 C_3 \cdot 1 \times \frac{5!}{2!} \\ & =\frac{4!}{2!} \times \frac{5 \times 4}{2 \times 1} \times \frac{5!}{2!} \\ & =12 \times 5 \times 120=7200 \end{aligned} $

Selecting 1 vowel 2 consonants (arrangement in which vowel is at middle place)

$ \begin{aligned} & =\left({ }^3 C_1 \cdot{ }^4 C_2\right) 2!+\left({ }^3 C_1 \cdot{ }^1 C_1\right) \cdot 1 \\ & =3 \times 6 \times 2+3=39 \end{aligned} $

Similarly,

Selecting, 2 vowels 1 consonant

$ ={ }^3 C_2 \cdot{ }^4 C_1(2!) \times 2=3 \times 4 \times 4=48 $

Selecting 3 vowels $=3!=6$

Total words $=39+48+6=93$

6 Boys and 4 girls

Number of arrangements in a row such that between any 2 girls there must be exactly 2 boys is

$ \begin{aligned} &=\xrightarrow{G_1}{} \uparrow \xrightarrow{G_2} \uparrow \xrightarrow{G_3} \uparrow{ }\xrightarrow{G_4}\\ & =4!\left(\frac{6!}{(2!)^3 \cdot 3!}\right) \cdot 3!(2!)^3 \\ & =6!\times 24=144(5!) \end{aligned} $

∵ Total number of ways in which train has stopped at exactly 5 stations $={ }^{15} C_5$

And total number of ways in which train has not stopped in consecutive stations

$ x_1+x_2+x_3+x_4+x_5+L=15-4=11 $

$x_i=$ number of station left before $i^{\text {th }}$ station

$ x_1 \geq 0 x_2, x_3, x_4, x_5 \geq 1 \text {, and } L=\text { station } $

left after 4 stations

$ \begin{aligned} x_1+\left(x_2-1\right)+ & \left(x_3-1\right)+\left(x_4-1\right) +\left(x_5-1\right)+L \\ = & 11-4=7 \end{aligned} $

$ \begin{aligned} \therefore \text { Total number of solution } & ={ }^{7+5-1} C_{6-1} \\ & ={ }^{11} C_5 \end{aligned} $

$ \text { ∴ Required ways }={ }^{15} C_5-{ }^{11} C_5 $

Number of all possible ways of distributing eight identical apples among 3 persons is number of non-negative integral solution of equation is

$ \begin{aligned} & x+y+z=8 \\ & \therefore{ }^{8+3-1} C_{3-1} \\ & \quad={ }^{10} C_2 \\ & \quad=\frac{10 \times 9}{2 \times 1}=45 \end{aligned} $

The number of words in CAB and NET both appears $=$ CAB NET $I=3$ !

∴ Required number of ways

$ \begin{aligned} & =7!-(5!+5!-3!) \\ & =5040-(120+120-6) \\ & =5040-234 \\ & =4806 \end{aligned} $

We have, $x+y+z+t=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When $x \geq 2, z \geq 5$

let $\quad x^{\prime}=x-2$

$\Rightarrow \quad x=x^{\prime}+2$

$ z^{\prime}=z-5 $

$\Rightarrow \quad z=z^{\prime}+5$

Here, $x^{\prime}, z^{\prime}, y \geq 0$ and $t \geq 0$ are non-negative integers.

$\therefore$ Substitude these value in Eq. (i), we get

$ \begin{aligned} & \left(x^{\prime}+2\right)+y+\left(z^{\prime}+5\right)+t=10 \\ \Rightarrow \quad & x^{\prime}+y+z^{\prime}+t=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

$\therefore$ Number of non-negative integral solution for Eq. (ii)

$ \begin{aligned} & ={ }^{3+4-1} C_{4-1}={ }^6 C_3 \\ & =\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20 \end{aligned} $

Integers lying between 1000 and 10000 such that the sum of all the digits in each of those numbers becomes 30 .

$\therefore$ Coefficient of $x^{30}$ in the expansion of

$ \begin{aligned} \left(x^1+x^2\right. & \left.+\ldots+x^9\right)\left(x^0+x^1+\ldots+x^9\right)^3 \\ & =x\left(\frac{1-x^9}{1-x}\right)\left(\frac{1-x^{10}}{1-x}\right)^3 \\ & =x\left(1-x^9\right)\left(1-x^{10}\right)^3(1-x)^{-4} \end{aligned} $

$\Rightarrow$ Coefficient of $x^{29}$ in

$ \left(1-x^9\right)\left(1-x^{10}\right)^3(1-x)^{-4} $

$ \begin{aligned} & =\left(1-x^9\right)\left(1-3 x^{10}+3 x^{20}-x^{30}\right)(1-x)^{-4} \\ & =\left(1-x^9-3 x^{10}+3 x^{19}+3 x^{20}-3 x^{29}\right. \\ & \left.\quad \quad-x^{30}+x^{39}\right)(1-x)^{-4} \end{aligned} $

Here, $(1-x)^{-4}=\sum_{k=0}^{\infty} k+{ }^3 C_3 x^k$

$\therefore$ Coefficient of $x^{29}$ from $1 \cdot(1-x)^{-4}$

$ ={ }^{29+3} C_3={ }^{32} C_3=4960 $

Coefficient of $x^{20}$ from $-x^9 \cdot(1-x)^{-4}$

$ =-{ }^{20+3} C_3=-{ }^{23} C_3=-1771 $

Coefficient of $x^{19}$ from $-3 x^{10} \cdot(1-x)^{-4}$

$ \begin{aligned} & =-3\left({ }^{19+3} C_3\right) \\ & =-3(1540)=-4620 \end{aligned} $

Coefficient of $x^{10}$ from $3 x^{19}(1-x)^{-4}$

(Means form is $3 x^{19} \Sigma^{k+3} C_3 x^k$ need

$ \begin{aligned} x^{29}, k & =10) \\ & =3^{10+{ }^3} C_3=3\left({ }^{13} C_3\right)=858 \end{aligned} $

Coefficient of $x^9$ from $3 x^{20} \cdot(1-x)^{-4}$

$ =3\left({ }^{9+3} C_3\right)=3(220)=660 $

Coefficient of $x^0$ from $-3 x^{29} \cdot(1-x)^{-4}$

$ =-3\left({ }^{0+3} C_3\right)=-3(1)=-3 $

$\therefore$ Total numbers

$ \begin{aligned} & =4960-1771-4620+858+660-3 \\ & =84 \end{aligned} $

Given word in most

Words start with $M=3!=6$

Words start with $\mathrm{O}=3!=6$

Words start with $\mathrm{SM}=2!=2$

Words start with $\mathrm{SO}=2!=2$

Words start with STM = 1 ! = 1

Words start with STO is STOM

Rank of STOM is $(6+6+2+2+1)+1 =18$

The total number of ways to choose any subset of the 5 options $=2^5=32$

These 32 subsets include choosing 2, 3, 4 or 5 options.

But choosing options 0 and 1 are not allowed.

Now, subsets with 0 options

$ ={ }^5 C_0=\frac{5!}{0!5!}=1 $

Subsets with 1 options

$ ={ }^5 C_1=\frac{5!}{1!4!}=5 $

So, total subsets with options 0 and 1

$ =1+5=6 $

Thus, the number of ways in which the student can answer that questions

$ =32-6=26 $

Let the triangle be $A B C$ with point $(x, y)$,

where $0 \leq x \leq 4$ and $0 \leq y \leq 4$

So, the points $(x, y)$ are $(0,0),(0,1)$, $(0,2),(0,3),(0,4),(1,0),(1,1),(1,2)$, $(1,3),(1,4),(2,0),(2,1),(2,2),(2,3)$, $(2,4),(3,0),(3,1),(3,2),(3,3),(3,4)$, $(4,0),(4,1),(4,2),(4,3),(4,4)$.

Total coordinates are 25 , but we need to select only 3 .

Now, the number of triangles with sets of 3 non-collinear points. Total number of ways to choose 3 points $={ }^{25} C_3$

But, we must subtract the number of collinear triplets.

Case I Horizontal lines

Number of ways to choose 3 collinear points

$ \Rightarrow \quad{ }^5 C_3=10 $

So, in 5 rows $=5 \times 10=50$

Case II Vertical lines

Same as above $\rightarrow 5$ columns

$ =5 \times 10=50 $

Now, diagonal from bottom-left to top-right

$=$ Length $3: 2$ such diagonals + Length

4 : 2 diagonals + Length 5:1 diagonal

$ \begin{aligned} = & 2\left({ }^3 C_3\right)+2\left({ }^4 C_3\right)+{ }^5 C_3 \\ & 2(1)+2(4)+10 \\ = & 2+8+10=20 \end{aligned} $

And, diagonals from top-left to bottom-right $=20$

Total collinear triplets

$ =50+50+20+20=140 $

So, total number of triangles $={ }^{25} C_3-140$

$ \Rightarrow \quad 2300-140=2160 $

Total Alphabets are A, D, E, H, L, N Letters before H are A, D, E.

So, the number of letters before $\mathrm{H}=3 \times 5$ !

$ =3 \times 120=360 $

Letters before $E$ are A, D.

So, the number of letters before $E=2 \times 4$ !

$ =2 \times 24=48 $

Letters before L are $\mathrm{A}, \mathrm{D}$.

So, the number of letters before $L$

$ =2 \times 3!=12 $

Letters before N is D

So, the number of letters with N only

$ \Rightarrow \quad 1 \times 1!=1 $

So, total words $=360+48+12+1=421$

Now, only 1 letter remaining, that is D .

So, Rank of HELAND $=421+1=422$

Calculate the number excluding the digit 5 .

(a) 1-digit number

$ =8(1,2,3,4,6,7,8,9) $

Similarly,

(c) 3-digit numbers $=8 \times 9 \times 9=648$

(d) 4-digit numbers $=8 \times 9 \times 9 \times 9=5832$

Total numbers between 1 and 10000 excluding the digit 5

$ =8+72+648+5832=6560 $

Hence, required numbers $=9999-6560$

$ =3439 $

Total number of people $=5+4=9$

So, total arrangements $=9$ !

To find $\alpha$ Let the particular man and woman be treated as a single unit or block when they are together.

Remaining 7 peoples +1 block $=8$ units

$ \therefore \quad \alpha=2 \times 8! $

and $\beta=9$ ! $-\alpha=9$ ! $-2 \times 8$ !

Hence, $\alpha: \beta=\frac{2 \times 8!}{8}: \frac{9!-2 \times 8!}{8!}$

$ =2: 7 $

Number of ways to choose 4 couples out of $4={ }^4 C_4=1$

From 4 couples, pick 1 person per couple $=2^4=16$

Team formula married couples $=2^2=4$

Hence, required ways $=16-4=12$

Total 8-digit numbers formed by $1,2,3,4,5,6,7,8$ is 8 !

Total 8 -digit numbers formed by $0,2,3$, $4,5,6,7,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,0,3$, $4,5,6,8,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,2,0$, $4,5,7,8,9$ is $7 \times 7$ !

Total 8-digit numbers formed by $0,1,2$, $3,5,6,7,8$ is $7 \times 7$ !

$ \begin{aligned} \text { Total } & =8!+28 \times 7! \\ & =(8+28) 7! \\ & =36 \times 7! \end{aligned} $

Given word, MATHEMATICS

$ \begin{array}{ll} 2 \mathrm{M} & 1 \mathrm{H} \\ 2 \mathrm{~A} & 1 \mathrm{E} \\ 2 \mathrm{~T} & 1 \mathrm{I} \\ & 1 \mathrm{C} \\ & 1 \mathrm{~S} \end{array} $

Number of words formed using 4 letters in which two are of same kind and 2 are different

$ \begin{aligned} & ={ }^3 C_1 \times{ }^7 C_2 \times \frac{4!}{2!} \\ & =3 \times 7 \times 3 \times 4 \times 3 \\ & =27 \times 28 \\ & =756 \end{aligned} $

Digits $-0,5,6,7,8$ and 9

No. greater than 6000

Case I 4 digit numbers greater than 6000

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=240 \\ & 4543 \end{aligned} $

$ \begin{aligned} &\text { Case II 5-digit numbers }\\ &\begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 55432 \end{aligned} \end{aligned} $

Case III 6-digit numbers

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 554321 \end{aligned} $

∴ Total $=1440$

∴ Required case

$=(2$ particular are in Ist group $)+$ (2 particular are in 3rd group) + (one is in Ist group and other is in 3rd group)

$ \begin{aligned} & =\frac{13!}{1!5!7!}+\frac{13!}{3!5!5!}+\frac{13!}{2!5!6!}(24) \\ & =\frac{13!}{7!5!}+\frac{13!}{6!5!}+\frac{13!}{5!6!}=\frac{13!(1+7+7)}{7!5!} \\ & =\frac{13 \times 12 \times 11!\times 15}{7!\times 5 \times 4 \times 3!}=\frac{117(11!)}{7!3!} \end{aligned} $

The digit are 1 to 9 . Each such number is strictly increasing sequence of digits, i.e., digits don't repeat and are in order.

e.g., 12, 135, 1239 and soon.

So, total such numbers in the number of increasing digit sequence of length 2 to 4.

We choose subsets of digits from 1 to 9 of lengths 2,3 and 4 in increasing order.

$\therefore 2$-digits numbers

$ ={ }^9 C_2=\frac{9!}{2!7!}=\frac{9 \times 8}{2}=36 $

3-digits numbers

$ ={ }^9 C_3=\frac{9!}{3!6!}=\frac{9 \times 8 \times 7 \times 6!}{6 \times 6!}=84 $

$ \begin{aligned} &\begin{aligned} 4 \text {-digits numbers } & ={ }^9 C_4=\frac{9!}{4!5!} \\ & =\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 5!}=126 \end{aligned}\\ &\text { So, total such digits }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=36+84+126=246 \end{aligned} $

Given, word is AGAIN,

So the letters in alphabatical order : A, A, G, I, N

Words starting with $A=\frac{4!}{2!}=12$

Words starting with $G=\frac{4!}{2!}=12$

Words starting with $I=\frac{4!}{2!}=12$

Words starting with $N=\frac{4!}{2!}=12$

So, total words $=12+12+12+12=48$

Thus, the 50th word starts with $N$.

49th word = NAAGI

$\therefore 50$ th word $=$ NAAIG.

We select exactly I wicket keeper from 4 .

So, number of ways $={ }^4 C_1=4$

Now, we have 10 players left to select from batsmen, bowlers and all-rounders with the condition.

Batsmen $\geq 4$, Bowlers $\geq 3$, All-rounders $\geq 2$

So, total of these three $=10$

So, number of batsmen, $B$ from 4 to 6 , number of bowlers, $L$ (say)from 3 to 6 and number of all-rounder, $A$ (say) from 2 to 4 .

$\therefore B+L+A=10$ (after 1 wicket keeper)

Now, the possible combinations for, (B, L, A) are

$(4,4,2),(4,3,3),(5,3,2)$

Total number of ways for 10 players

$ \begin{aligned} & =\left({ }^6 C_4{ }^6 C_4{ }^4 C_2+{ }^6 C_4{ }^6 C_3{ }^4 C_3+{ }^6 C_5{ }^6 C_3{ }^4 C_2\right) \\ & =(15 \times 15 \times 6+15 \times 20 \times 4+6 \times 20 \times 6) \\ & =(1350+1200+720)=3270 \end{aligned} $

$\therefore$ Total number of ways for 11 players

$ \begin{aligned} & =4 \times 3270 \\ & =13080 \end{aligned} $

Number of combinations of choosing 4 digits out of 4

$ ={ }^5 C_4=5 $

Each 4-digit set can be arranged in 4! ways $=24$ ways

So, total number of 4 -digit numbers

$ =5 \times 24=120 $

In each group, each digit appears

$ \frac{24}{4}=6 \text { times } $

So, contribution per group

$ =\text { sum of digits } \times 6 \times(1000+100+10+1) $

Now, compute total sum of digits for all 5 groups

$ \begin{array}{r} \begin{array}{r} (2+3+5+8)+(1+3+5+8) \\ +(1+2+5+8)+(1+2+3+8) \\ +(1+2+3+5) \end{array} \\ =18+17+16+14+11=76 \end{array} $

$ \text { Hence, total sum } \begin{aligned} & =76 \times 6666 \\ & =506616 \end{aligned} $

Total possible ordered pairs $(p, q)$ of first 50 natural numbers

$ =50 \times 50=2500 $

Out of these, exactly half will satisfy $p>q$ and there are 50 pairs where $p=q \{(1,1),(2,2) \ldots(50,50)\}$

So, number of pairs with $p \neq q$

$ =2500-50=2450 $

and for $p>q$

$ =\frac{2450}{2}=1225 $

Fathers $(F)=5$, Teachers $(T)=6$, students $(S)=4$

for, $T+F+S=7, T \geq 0$, thus $F \geq 1, S \geq 1$ total combination ( $T, F, S$ )

$ (4,1,2),(4,2,1),(5,1,1),(3,2,2) $

Case $\mathbf{I}(T=4, F=1, S=2)$

$ \begin{aligned} & ={ }^6 C_4 \cdot{ }^5 C_1 \cdot{ }^4 C_2 \\ & =15 \times 5 \times 6=450 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{Case \,I I}\,\,(T & =4, F=2, S=1) \\ & ={ }^6 C_4 \cdot{ }^5 C_2 \cdot{ }^4 C_1 \\ & =15 \times 10 \times 4=600 \end{aligned} $

Case III ( $T=5, F=1, S=1$ )

$ \begin{aligned} & ={ }^6 C_5 \cdot{ }^5 C_1 \cdot{ }^4 C_1 \\ & =6 \times 5 \times 4=120 \end{aligned} $

Case IV ( $T=3, F=2, S=2$ )

$ \begin{aligned} & ={ }^6 C_3 \cdot{ }^5 C_2 \cdot{ }^4 C_2 \\ & =20 \times 10 \times 6=1200 \end{aligned} $

Hence, total number of ways

$ \begin{aligned} & =450+600+120+1200 \\ & =2370 \end{aligned} $

The total number of ways that all sisters and brothers seated around a circle is 10 !

The number of ways that the three sisters seated together along with the brothers is $8!\times 3$ !

So, required number of ways

$ =10!-8!3!=8!(10 \times 9-6)=8!\times 84 $

Given, 8 students in a class room 4 of them are chosen and arrange in ground table and 4 in row

$ \begin{aligned} & \therefore \text { Total arrangement }={ }^8 C_4(3!+4!) \\ & \quad=\frac{8!}{4!4!}(6+24)=70 \times 30 \\ & \quad=2100 \end{aligned} $

Give word VARIABLE

A twice and other are different arrangement of three letter words.

Case 12 alike 1 distinct

$ ={ }^1 C_1 \times{ }^6 C_1 \times \frac{3!}{2!}=6 \times 3=18 $

Case II All are distinct

$ \begin{aligned} { }^7 C_3 \times 3! & =35 \times 6=210 \\ \therefore \text { Total } & =210+18=288 \\ n(S) & =288 \end{aligned} $

Middle letter is consonant.

$ =6 \times 4 \times 5=120 $

$ =1 \times 4=4 $

Total favourable case $=120+4=124$

$\therefore$ Required probability $=\frac{124}{288}=\frac{31}{57}$

$ \begin{aligned} &\\ &\begin{aligned} & \text { }{ }^{27} P_{r+7}=7722{ }^{25} P_{r+4} \\ & \Rightarrow \frac{27!}{20-r!}=7722 \cdot \frac{25!}{21-r!} \\ & \Rightarrow \frac{27 \cdot 26 \cdot 25!}{20-r!}=7722 \cdot \frac{25!}{(21-r) 20-r!} \\ & \Rightarrow 27 \cdot 26=7722 \times \frac{1}{(21-r)} \\ & \Rightarrow \text { By option put } r=10 \\ & \Rightarrow \quad 702=702 \end{aligned} \end{aligned} $

Let number of side $=n$

The number of diagonals of a regular

$ \begin{array}{ll} \Rightarrow & \frac{n(n-3)}{2}=35 \\ \Rightarrow & n^2-3 n=70 \\ \Rightarrow & n^2-3 n-70=0 \\ \Rightarrow & n^2-10 n+7 n-70=0 \\ \Rightarrow & n(n-10)+7(n-10)=0 \\ \Rightarrow & (n-10)(n+7)=0 \\ \Rightarrow & n=10, n=-7(\text { neglected }) \\ \therefore & n=10 \end{array} $

ASSIGNMENT

Total $=10$

2 S and 2 N

Different $=8$

A, I, G, M, E, T

Case I All are different

$ ={ }^8 C_4 \times 4!=1680 $

Case II Two are same and two are different

$ \begin{aligned} & ={ }^2 C_1 \times{ }^7 C_2 \times \frac{4!}{2!}=756 \\ & =2!\times \frac{7!}{2!\cdot 5!} \times \frac{4!}{2!} \\ & =\frac{7 \times 6 \times 5!}{2!\times 5!} \times 4 \times 3 \times 2!=504 \end{aligned} $

Case III Two are same and two are same

$ ={ }^2 C_2 \times \frac{4!}{2!\times 2!}=6 $

Total $=1680+504+6=2190$

We have, $\mathrm{E}, \mathrm{E}, \mathrm{L}, \mathrm{R}, \mathrm{T}, \mathrm{T}$

Words starting with $\mathrm{E}, \mathrm{L}, \mathrm{R}$, where first letter is T

(i) Starting with $\mathrm{E}: \frac{5!}{2!}=60$

(ii) Starting with $\mathrm{L}: \frac{5!}{2!2!}=30$

(iii) Starting with $\mathrm{R}: \frac{5!}{2!2!}=30$

Total $=60+30+30=120$

Fixed TE next is T

Letters before T is E, L, R

Each gives $3!=6$

$ \Rightarrow 3 \times 6=18 $

Now, fixed TETL, next is E one word before TETLE

$k$ is remaining letter which is before 2 letters.

Final letter is TETLER =1

Rank $=120+18+2+1=141$.

Given digits $0,1,2,3,5,7$

To find rank of 70513

Total number start with $1=5!=120$

Number start with $2=5!=120$

Number start with $3=5!=120$

Number start with $5=5!=120$

Number start with $701=6$

Number start with $702=6$

Number start with $703=6$

$ \begin{aligned} & 70512=1 \\ & 70513=1 \end{aligned} $

Rank of 70513

$ \begin{aligned} & =4 \times 120+3 \times 6+2 \\ & =500 \end{aligned} $

Number of divisor of 7!.

$ \begin{aligned} 7! & =7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ & =2^4 \times 3^2 \times 5^1 \times 7^1 \end{aligned} $

Number of divisor

$ \begin{aligned} & =(4+1)(2+1)(1+1)(1+1) \\ & =60 \end{aligned} $

COMBINATION

Vowels- O, O, I, I, A = 5

$ \begin{aligned} &=4 \times \frac{8!}{2!2!} \times 2(\mathrm{C} \text { and } \mathrm{N} \text { are also be arranged })\\ &=2 \times 8! \end{aligned} $

Total number of fruits $=3 \times 12=36$

These fruits distributed among 9 persons equals $\because$ Order is taken'.

Total number of ways $=\frac{36!}{(4!)^9 9!} \times 9!=\frac{36!}{(4!)^9}$.

We have, $\sum\limits_{i=0}^m{ }^{10} C_i{ }^{20} C_{m-i}$

$ ={ }^{10} C_0{ }^{20} C_m+{ }^{10} C_1{ }^{20} C_{m-1}+{ }^{10} C_2{ }^{20} C_{m-2}+\ldots+{ }^{10} C_m \cdot{ }^{20} C_0 $

Coefficient of $x^m$ in the expansion of the product

$ (1+x)^{10}(1+x)^{20} $

$=$ Coefficient of $x^m$ in $(1+x)^{30}$

$ ={ }^{30} C_m . $

Maximum value of ${ }^{30} C_m={ }^{30} C_{15}$

$ \therefore \quad m=15 $

We have, $x y z=30$

$30=1,2,3,5,6,10,15,30$ are the integral factor of 30

$ \begin{aligned} & 30=3 \times 2 \times 5=3! \\ & 30=6 \times 5 \times 1=3! \\ & 30=10 \times 3 \times 1=3! \\ & 30=15 \times 2 \times 1=3! \\ & 30=30 \times 1 \times 1=\frac{3!}{2!} \end{aligned} $

$ \begin{aligned} \therefore \text { Total number of factor } & =4 \times 3!+\frac{3!}{2!} \\ & =24+3=27 \end{aligned} $

Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are

$ \begin{aligned} & \mathrm{I}-2 \\ & \mathrm{~N}-4 \\ & \mathrm{C}-2 \\ & \mathrm{O}-1 \\ & \mathrm{~V}-1 \\ & \mathrm{E}-3 \end{aligned} $

Total number of 5-letters words

(i) When 5 distinct letters, number of words $=6 P_5=720$

(ii) One pair and 3 distinct letters,

$ \begin{aligned} \text { number of words } & ={ }^4 C_1 \times{ }^5 C_3 \times \frac{5!}{2!} \\ & =4 \times 10 \times 60=2400 \end{aligned} $

(iii) One triple and 2 distinct letters, number of words

$ ={ }^2 C_1 \times{ }^5 C_2 \times \frac{5!}{3!} $

$ =2 \times 10 \times 20=400 $

(iv) Two pairs and one distinct, number

$ \begin{aligned} \text { of words } & ={ }^4 C_2 \times{ }^4 C_1 \times \frac{5!}{2!2!} \\ & =6 \times 4 \times 30=720 \end{aligned} $

(v) One triplet and one pair, number of words

$ =4 \times \frac{5!}{3!2!}=4 \times 10=40 $

(vi) One 4 a like and one different

$ ={ }^1 C_1 \times{ }^5 C_1 \times \frac{5!}{4!}=25 $

$\therefore$ Number of words with at least one repealed letters

$ \begin{aligned} & =(720+2400+400+720+40+25)-720 \\ & =3585 \end{aligned} $

Given, PERFECTION

Vowels E E I O and consonants PRFC T M.

Vowels can be arranged by $\frac{4!}{2!}$

$ \begin{aligned} & =4 \cdot 3!=2!3! \\ & E \times X E \times X \times X O \end{aligned} $

Consonants are arranged at $\times$ places by 6 ! ways.

$\because$ Total number of ways $=6!\times 3!\times 2!$