Permutations and Combinations

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 $=6!\times 3!\times 2!$

$\begin{aligned} & 2 M \\ & 2 A \\ & 2 T \\ & H, E, I, C, S \end{aligned}$

Case-I

2 Alike 2 Alike 1 Diff

${ }^3 C_2 \times{ }^6 C_1=18$

Case-II

2 Alike + 3 Diff

${ }^3 C_1 \times{ }^7 C_3=105$

Case-III

All different

${ }^8 C_5=56$

Total ways $=179$

$\begin{aligned} & A=\{2,3,5,7,11\} \\ & f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right] \\ & \text { Ranges } f(x)=\{2,3,5,6,8\} \end{aligned}$

$\text { Number of one-one } A \rightarrow R_f$

$5 \times 4 \times 3 \times 2 \times 1=120$

NAGPUR

Word at $315^{\text {th }}$ position

$\begin{aligned} & \text { A...... }=5!=120 \\ & \text { G....... }=5!=120 \\ & \text { NA..... }=4!=24 \\ & \text { NG..... }=4!=24 \\ & \text { NP..... }=4!=24 \end{aligned}$

..... Till 312 words

$313^{\text {th }} \text { word }=\text { NRAGPU }$

$314^{\text {th }}$ word $=$ $\mathrm{NRAGUP}$

$315^{\text {th }} \text { word }=\text { NRAPGU }$

Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.

Step-by-Step Solution:

1. Write the given proportions involving binomial coefficients:

$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \frac{r}{n} \times \binom{n}{r} = 55 : 35 : 21 $

1. Simplify the proportions:

$ \frac{n+1}{r+1} = \frac{55}{35} \quad \text{and} \quad \frac{n}{r} = \frac{35}{21} $

1. From the simplified ratios, we establish the following two equations:

$ \frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)} $

$ \frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)} $

1. Solve equations (1) and (2) simultaneously:

• From equation (2), solve for $n$:

$ 3n = 5r \quad \Rightarrow \quad n = \frac{5r}{3} $

• Substitute $n$ into equation (1):

$ 7 \left(\frac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4 $

$ \frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6 $

• Substituting $r$ back to find $n$:

$ n = \frac{5 \times 6}{3} = 10 $

1. Compute $2n + 5r$:

$ 2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50 $

Thus, the value of $2n + 5r$ is 50.

To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.

Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon. The number of ways to choose 3 vertices out of 8 is given by the combination formula:

$ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $

Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon. Let's analyze how many such invalid triangles there can be.

Consider each side of the octagon. For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side). Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon. Hence, for each side of the octagon, there are 5 such triangles.

Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:

$ 8 \times 5 = 40 $

Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:

$ 56 - 40 = 16 $

So, the correct answer is:

Option B: 16

Given set $S=\left\{2^1, 2^2, \ldots 2^9\right\}$ which consist of 9 elements.

Maximum number of possible partitions (in set $A, B$ and $C$)

$={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680$

To find the $50^{\text{th}}$ word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.

First, calculate the total number of permutations of these letters:

$ \frac{5!}{2!} = 60 $

Let's arrange the letters in alphabetical order first: B, B, H, J, O.

We need to systematically count the words while following dictionary order:

1. Words starting with B:

• Next position letters: B, H, J, O
• Number of permutations: $\frac{4!}{1!} = 24$ words

Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.

2. Words starting with H:

• Next position letters: B, B, J, O
• Number of permutations: $\frac{4!}{2!} = 12$ words

After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.

3. Words starting with J:

• Next position letters: B, B, H, O
• Number of permutations: $\frac{4!}{2!} = 12$ words

Now tally is 36 + 12 = 48 words.

Still 2 more to reach 50.

4. Words starting with O:

• Next position letters: B, B, H, J
• Number of permutations: $\frac{4!}{2!} = 12$ words
• Our answer must be here since 48 + 2 more = 50 total.

First permutation: $OBB H J$

Second permutation (50th word): $OBB J H$

So, the $50^{\text{th}}$ word is: OBBJH

Thus, the correct answer is Option A: OBBJH.

Number of points on side $A B=5$

Number of points on side $B C=6$

Number of points on side $A C=7$

Number of ways selecting three points from side

$A B={ }^5 C_3$

Number of ways selecting three points from side

$B C={ }^6 C_3$

Number of ways selecting three points from side

$A C={ }^7 C_3$

Total number of triangle possible formed using the points $P_1 P_2 \ldots P_{18}$

$\begin{aligned} & ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\\\ & =816-10-20-35 \\\\ & =751 \end{aligned}$

To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.

The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.

The partitions of 5 into at most 4 parts are as follows :

1. $(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.


2. $(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.


3. $(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.


4. $(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.


5. $(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.


6. $(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.

However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts. This means that partitions like $(5, 0, 0, 0)$, $(0, 5, 0, 0)$, $(0, 0, 5, 0)$, and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.

So we end up with 6 distinct partition scenarios.

Now, for each partition, we need to find the number of ways to assign the employees according to each partition :

1. $(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.


2. $(4, 1, 0, 0)$ - For the group with 4 employees, there are ${}^5{C_4}$ ways to choose which 4 out of the 5 will be together. Then the remaining employee is automatically assigned to the second office. So, there are ${}^5{C_4}$ ways for this partition.


3. $(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are ${}^2{C_2}$ ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped). But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2. So, there are ${}^5{C_3}$ ways for this partition.


4. $(3, 1, 1, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways. Then we have two indistinguishable offices, each with 1 employee. The order does not matter as offices are indistinguishable. So, there are simply ${}^5{C_3}$ ways.


5. $(2, 2, 1, 0)$ – For the first group of 2 employees, there are ${}^5{C_2}$ ways to choose them. For the next group of 2 employees, there are ${}^3{C_2}$ ways from the remaining 3. Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider. However, we have double-counted since the two groups of two are indistinguishable. To adjust for this, we divide by 2. This gives us $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2}$ ways.


6. $(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them. The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.

Let's compute these cases :

• $(5, 0, 0, 0)$ corresponds to $1$ way.


• $(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.


• $(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.


• $(2, 2, 1, 0)$ corresponds to $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \frac{10 \cdot 3}{2} = 15$ ways.


• $(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.

Adding up all the ways we get :

$ 1 + 5 + 10 + 10 + 15 + 10 = 51 $

Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\mathrm{n}$) is 51, which corresponds to Option C.