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**,**2 and 3 only, is

(A) 55 (B) 66 (C) 77 (D) 88

**[JEE 2009, 3]**

**Sol.**(C)

For sum to be 10

Case-I $\quad 1,2,3,1,1,1,1,=\frac{7 !}{5 !}$

Case-II $\quad 2,2,2,1,1,1,1,=\frac{7 !}{3 ! 4 !}$

So total number $=77$

(A) 25 (B) 34 (C) 42 (D) 41

**[JEE 10, 5M, –2M]**

**Sol.**(D)

Case- $\mathrm{I}:$ The number of elements in the pairs $(\mathrm{A}, \mathrm{B})$ can be $(1,1) ;(1,2) ;(1,3) ;(2,2)$

$=^{4} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \times^{3} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \times^{3} \mathrm{C}_{3}+\frac{^{4} \mathrm{C}_{2} \cdot^{2} \mathrm{C}_{2}}{2}=25$

Case- $\mathrm{II}:$ Number of pairs with $\phi$ as one of subsets $=2^{4}=16$

$\therefore$ Total pairs $=25+16=41$

(A) 75 (B) 150 (C) 210 (D) 243

**[JEE 2012, 3M, –1M]**

**Sol.**(B)

Balls can be distributed as $1,1,3$ or $1,2,2$ to each person.

When $1,1,3$ balls are distributed to each person, then total number of ways:

$=\frac{5 !}{1 ! 1 ! 3 !} \cdot \frac{1}{2 !} 3 !=60$

When $1,2,2$ balls are distributed to each person, then total number of ways :

$=\frac{5 !}{1 ! 2 ! 2 !} \cdot \frac{1}{2 !} 3 !=90$

$\therefore$ total $=60+90=150$

Let $\mathrm{a}_{\mathrm{n}}$ denotes the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let $b_{n}$ = the number of such n-digit integers ending with digit 1 and $\mathrm{C}_{\mathrm{n}}$ = the number of such n-digit integers ending with digit 0.

(A) 7 (B) 8 (C) 9 (D) 11

**[JEE 2012, 3M, –1M] **

**Sol.**(B)

$\mathrm{a}_{1}=1, \mathrm{a}_{2}=2$

So $a_{3}=3, \quad a_{4}=5 a_{5}=8$

$\Rightarrow b_{6}=a_{5}=8$

(A) $a_{17}=a_{16}+a_{15}$

(B) $\mathrm{c}_{17} \neq \mathrm{c}_{16}+\mathrm{c}_{15}$

(C) $\mathrm{b}_{17} \neq \mathrm{b}_{16}+\mathrm{c}_{16}$

(D) $a_{17}=c_{17}+b_{16}$

**[JEE 2012, 3M, –1M]**

**Sol.**(A)

$a_{n}=a_{n-1}+a_{n-2}$

put $n=17$

$a_{17}=a_{16}+a_{15} \quad($ A) is correct

$c_{n}=c_{n-1}+c_{n-2}$

$\mathrm{c}_{\mathrm{n}}=\mathrm{c}_{\mathrm{n}-1}+\mathrm{c}_{\mathrm{n}-2}$

$\mathrm{So}$ put $\mathrm{n}=17$

$\mathrm{c}_{17}=\mathrm{c}_{16}+\mathrm{c}_{15} \quad(\mathrm{B})$ is incorrect

$\mathrm{b}_{\mathrm{n}}=\mathrm{b}_{\mathrm{n}-1}+\mathrm{b}_{\mathrm{n}-2}$

put $\mathrm{n}=17$

$\mathrm{b}_{17}=\mathrm{b}_{16}+\mathrm{b}_{15} \quad(\mathrm{C})$ is incorrect

$\mathrm{a}_{17}=\mathrm{a}_{16}+\mathrm{a}_{15}$

while $(\mathrm{D})$ says $\mathrm{a}_{17}=\mathrm{a}_{15}+\mathrm{a}_{15} \quad$ (D) is incorrect

**[JEE(Advanced)-2014, 3]**

**Sol.**5

as $n_{1} \geq 1, n_{2} \geq 2, n_{3} \geq 3, n_{4} \geq 4, n_{5} \geq 5$

Let $n_{1}-1=x_{1} \geq 0, n_{2}-2=x_{2} \geq 0 \ldots \ldots \ldots \ldots n_{5}-5=x_{5} \geq 0$

$\Rightarrow$ New equation will be

$\quad x_{1}+1+x_{2}+2+\ldots \ldots x_{5}+5=20$

$\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=20-15=5$

Now $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$

AND

Numbr of red line segments $=^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$

Number of blue line segments $=\mathrm{n}$

$\therefore^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}=\mathrm{n}$

$\frac{\mathrm{n}(\mathrm{n}-1)}{2}=2 \mathrm{n} \Rightarrow \mathrm{n}=5$ Ans.

**[JEE(Advanced)-2014, 3]**

**Sol.**7

(A) 264 (B) 265 (C) 53 (D) 67

**[JEE(Advanced)-2014, 3(–1)]**

**Sol.**(C)

Total number of dearrangement

6! $\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}\right]$

$=360-120+30-6+1$

$=240+25=265$

There are equal chances that card 1 goes into any envelope from 2 to $6 \quad \therefore \quad \frac{1}{5}(265)=53$

**[JEE (Advanced) 2015, 4M, –0M]**

**Sol.**5

$n=6 ! \times 5 !$

$n=6 ! \times 4 ! \times 5 \times^{5} C_{4}$

$\frac{m}{n}=\frac{6 ! \times 4 ! \times 5 \times^{5} C_{4}}{6 ! \times 5 !}=5$

(A) 380 (B) 320 (C) 260 (D) 95

**[JEE (Advanced) 2016]**

**Sol.**(A)

**[JEE (Advanced) 2017]**

**Sol.**5

$\mathrm{x}=10 !$

$\mathrm{y}=^{10} \mathrm{C}_{1}^{9} \mathrm{C}_{8} \frac{10 !}{2 !}$

Then $\mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=$b

(A) 125 (B) 252 (C) 210 (D) 126

**[JEE (Advanced) 2017]**

**Sol.**(D)

$\mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=$ Total ways $-\{\text { when no odd }\}$

Total ways $=^{9} \mathrm{C}_{5}$

Number of ways when no odd, is zero $\quad(\because \text { only available even are } 2,4,6,8)$

$\therefore$ Ans $:^{9} \mathrm{C}_{5}-$ zero $=126$

**[JEE (Advanced) 2018,3(0)]**

**Sol.**625

Option for last two digits are $(12),(24),(32),(44)$ are $(52) .$

$\therefore \quad$ Total No. of digits

$=5 \times 5 \times 5 \times 5=625$

(i) Let $\alpha_{1}$ be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 body and 2 girls.

(ii) Let $\alpha_{2}$ be the total number of ways in which the committe can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.

(iii) Let $\alpha_{3}$ be the total number of ways in which the committe can be formed such that the committee has 5 members, at least 2 of them being girls.

(iv) Let $\alpha_{4}$ be the total number of ways in which the committee can be formed such that the commitee has 4 members, having at least 2 girls and such that both $\mathrm{M}_{1}$ and $\mathrm{G}_{1}$ are **NOT** in the committee together.

**[JEE (Advanced) 2018,3(–1)]**

**Sol.**(C)

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