JEE Advanced Maths – Permutations and Combinations PYQ

Q 1 :

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is [2016]

380
320
260
95

1

2

3

4

(1)

Either one boy will be selected or no boy will be selected.

Also out of four members one captain is to be selected.

$∴$ $Required number of ways = ({}^{4}C_{1} \times {}^{6}C_{3} + {}^{6}C_{4}) \times {}^{4}C_{1}$

$= (80 + 15) \times 4 = 380$

Q 2 :

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is [2012]

75
150
210
243

1

2

3

4

(2)

$∵$ Each person gets at least one ball.

$∴$ 3 persons can have 5 balls as follows.

Person	No. of balls	No. of balls
I	1	1
II	1	2
III	3	2

The number of ways to distribute balls 1,1,3 in first to three persons $= {}^{5}C_{1} \times {}^{4}C_{1} \times {}^{3}C_{3}$

Also 3 persons having 1,1 and 3 balls can be arranged in $\frac{3!}{2!}$ ways.

$∴$ Total number of ways to distribute 1,1,3 balls to the three persons $= {}^{5}C_{1} \times {}^{4}C_{1} \times {}^{3}C_{3} \times \frac{3!}{2!} = 60$

Similarly, total number of ways to distribute 1, 2, 2 balls to three persons $= {}^{5}C_{1} \times {}^{4}C_{2} \times {}^{2}C_{2} \times \frac{3!}{2!} = 90$

$∴$ $The required number of ways = 60 + 90 = 150$

Q 3 :

A rectangle with sides of length ( $2 m - 1$ ) and ( $2 n - 1$ ) units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is [2005]

[IMAGE 27]

${(m + n - 1)}^{2}$
$4^{m + n - 1}$
$m^{2} n^{2}$
$m (m + 1) n (n + 1)$

1

2

3

4

(3)

[IMAGE 28]

If we see the blocks in terms of lines then there are $2 m$ vertical lines and $2 n$ horizontal lines. To form the required rectangle, we must select two horizontal lines, one even numbered (out of $2, 4, \dots, 2 n$ ) and one odd numbered (out of $1, 3, \dots, 2 n - 1$ ) and similarly two vertical lines.

The number of rectangles $= {}^{m}C_{1} \cdot {}^{m}C_{1} \cdot {}^{n}C_{1} \cdot {}^{n}C_{1} = m^{2} n^{2}$

Q 4 :

Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals [2001]

5
7
6
4

1

2

3

4

(2)

$T_{n} = {}^{n}C_{3}; T_{n + 1} = {}^{n + 1}C_{3}$

$Now, T_{n + 1} - T_{n} = 21 \Rightarrow {}^{n + 1}C_{3} - {}^{n}C_{3} = 21$

$\Rightarrow \frac{(n + 1) n (n - 1)}{3 \cdot 2 \cdot 1} - \frac{n (n - 1) (n - 2)}{3 \cdot 2 \cdot 1} = 21$

$\Rightarrow n (n - 1) (n + 1 - (n - 2)) = 126$

$\Rightarrow n (n - 1) = 42$

$\Rightarrow n (n - 1) = 7 \times 6 \Rightarrow n = 7$

Q 5 :

A group of 9 students, $s_{1}, s_{2}, \dots, s_{9}$ is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that $s_{1}$ cannot be selected for the team X, and $s_{2}$ cannot be selected for the team Y. Then the number of ways to form such teams is ________. [2024]

(665)

Number of required ways

$= \frac{9!}{2! 3! 4!} - (n (s_{1} \in X) + n (s_{2} \in Y) - n (s_{1} \in X and s_{2} \in Y))$

$= \frac{9!}{2! 3! 4!} - (\frac{8!}{1! 3! 4!} + \frac{8!}{2! 2! 4!} - \frac{7!}{1! 2! 4!}) = 665$

Q 6 :

Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{y}{9 x} =$ [2017]

(5)

$x = 10! and y = {}^{10}C_{1} \times {}^{9}C_{8} \times \frac{10!}{2!} = 10 \times 9 \times \frac{10!}{2!}$

$∴$ $\frac{y}{9 x} = \frac{10 \times 9 \times \frac{10!}{2}}{9 \times 10!} = 5$

Q 7 :

Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{m}{n}$ is [2015]

(5)

$Here,_B_{1}_B_{2}_B_{3}_B_{4}_B_{5}$

Out of 5 girls, 4 girls are together and 1 girl is separate. Now, to select 2 positions out of 6 positions between boys $= {}^{6}C_{2} ...(i)$

4 girls are to be selected out of 5 $= {}^{5}C_{4} ...(ii)$

Now, 2 groups of girls can be arranged in $2!$ ways $. ..(iii)$

Also, the group of 4 girls and 5 boys is arranged in $4! \times 5!$ ways $...(iv)$

Now, total number of ways $= {}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!$ $[from Eqs. (i), (ii), (iii) and (iv)]$

$∴$ $m = {}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!$ $and n = 5! \times 6!$

$\Rightarrow \frac{m}{n} = \frac{{}^{6}C_{2} \times {}^{5}C_{4} \times 2! \times 4! \times 5!}{6! \times 5!} = \frac{15 \times 5 \times 2 \times 4!}{6 \times 5 \times 4!} = 5$

Q 8 :

Let $n \geq 2$ be an integer. Take $n$ distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of $n$ is [2014]

(5)

Number of adjacent lines $= n$

Number of non-adjacent lines $= {}^{n}C_{2} - n$

$∴$ ${}^{n}C_{2} - n = n \Rightarrow \frac{n (n - 1)}{2} - 2 n = 0$

$\Rightarrow n^{2} - 5 n = 0 \Rightarrow n = 0 or 5$ $But n \geq 2 \Rightarrow n = 5$

Q 9 :

Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in $S$ , but 0210222 is NOT in $S$ .

Then the number of elements $x$ in $S$ such that at least one of the digits 0 and 1 appears exactly twice in $x$ , is equal to ______. [2025]

(762)

Let $A \to$ “0” appear exactly twice, and $B \to$ “1” appear exactly twice.

$∴$ $A \cap B \to “0” and “1” both appear exactly twice.$

[IMAGE 29]----------

$= \underset{(placing zero)}{{}^{6}C_{2}} \cdot (1) \cdot 2^{5} = \frac{6 \times 5}{2} \times 2^{5} = 480$

$For n (B)$

$Case 1: 1 at first place$

[IMAGE 30]----------

$Number of ways = \underset{(placing 1)}{{}^{6}C_{1}} \cdot (1) \cdot 2^{5} = 192$

$Case 2: 2 at first place$

[IMAGE 31]----------

$Number of ways = \underset{(placing 1)}{{}^{6}C_{2}} \cdot (1) \cdot 2^{4} = \frac{6 \times 5}{2} \times 2^{4} = 240$

$n (B) = 240 + 192$

$For n (A \cap B)$

[IMAGE 32]----------

$For n (A \cap B)$

$= \underset{placing zero}{{}^{6}C_{2}} \cdot (1) \times \underset{placing 1}{{}^{5}C_{2}} \cdot (1) \times \underset{2 at rest places 1}{(1)}$

$= \frac{6 \times 5}{2} \times \frac{5 \times 4}{2} = 150$

$∴$ $n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$= 480 + (192 + 240) - 150 = 762$

Q 10 :

An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1–15 June 2021 is ______. [2020]

(495)

We know that total number of ways of selection of $r$ days out of $n$ days such that no two of them are consecutive $= {}^{n - r + 1}C_{r}$

$∴$ Selection of 4 days out of 15 days such that no two of them are consecutive $= {}^{15 - 4 + 1}C_{4} = {}^{12}C_{4}$

$= \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} = 11 \times 5 \times 9 = 495$

Q 11 :

Let $S_{1} = {(i, j, k) : i, j, k \in {1, 2, \dots, 10}}$

$S_{2} = {(i, j) : 1 \leq i < j + 2 \leq 10, i, j \in {1, 2, \dots, 10}}$

$S_{3} = {(i, j, k, l) : 1 \leq i < j < k < l, i, j, k, l \in {1, 2, \dots, 10}}$

and $S_{4} = {(i, j, k, l) : i, j, k and l are distinct elements in {1, 2, \dots, 10}}$ .

If the total number of elements in the set $S_{r}$ is $n_{r}$ , $r = 1, 2, 3, 4$ , then which of the following statements is (are) TRUE? [2021]

$n_{1} = 1000$
$n_{2} = 44$
$n_{3} = 220$
$\frac{n_{4}}{12} = 420$

1

2

3

4

Select one or more options

(1, 2, 4)

Number of elements in $S_{1} = 10 \times 10 \times 10 = 1000$

Number of elements in $S_{2} = 9 (J = 8) + 8 (J = 7) + 7 (J = 6) + 6 (J = 5) + 5 (J = 4) + 4 (J = 3) + 3 (J = 2) + 2 (J = 1) = 44$

Number of elements in $S_{3} = {}^{10}C_{4} = 210$

Number of elements in $S_{4} = {}^{10}P_{4} = 210 \times 4! = 5040$

So, options (1), (2), (4) are correct.

Q 12 :

Let $S = {1, 2, 3, \dots, 9}$ . For $k = 1, 2, \dots, 5$ , let $N_{k}$ be the number of subsets of $S$ , each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$ [2017]

210
252
125
126

1

2

3

4

(4)

Here set $S$ contain 5 odd and 4 even numbers. Since each of $N_{k}$ contains five elements out of which exactly are odd.

$∴$ $N_{1} = {}^{5}C_{1} \times {}^{4}C_{4} = 5$

$N_{2} = {}^{5}C_{2} \times {}^{4}C_{3} = 40$

$N_{3} = {}^{5}C_{3} \times {}^{4}C_{2} = 60$

$N_{4} = {}^{5}C_{4} \times {}^{4}C_{1} = 20$

$N_{5} = {}^{5}C_{5} = 1$

$∴$ $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} = 126$

Q 13 :

For non-negative integers $s$ and $r$ , let

$(\binom{s}{r}) =$ ${\begin{matrix} \frac{s!}{r! (s - r)!} & if r \leq s, \\ 0 & if r > s . \end{matrix}$

For positive integers $m$ and $n$ , let

$g (m, n) = \sum_{p = 0}^{m + n} \frac{f (m, n, p)}{(\binom{n + p}{p})}$

where for any non-negative integer $p$ ,

$f (m, n, p) = \sum_{i = 0}^{p} (\binom{m}{i}) (\binom{n + i}{p}) (\binom{p + n}{p - i}) .$

Then which of the following statements is/are TRUE? [2020]

$g (m, n) = g (n, m) for all positive integers m, n,$
$g (m, n + 1) = g (m + 1, n) for all positive integers m, n$
$g (2 m, 2 n) = 2 g (m, n) for all positive integers m, n$
$g (2 m, 2 n) = (g {(m, n))}^{2} for all positive integers m, n$

1

2

3

4

Select one or more options

(1, 2, 4)

[IMAGE 33]

So, options (1), (2) and (4) are true.

Q 14 :

In a high school, a committee has to be formed from a group of 6 boys $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}$ and 5 girls $G_{1}, G_{2}, G_{3}, G_{4}, G_{5}$ .

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

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

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

(iv) Let $α_{4}$ be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both $M_{1}$ and $G_{1}$ are NOT in the committee together.

LIST-I LIST-II

P. The value of $α_{1}$ is 1. 136

Q. The value of $α_{2}$ is 2. 189

R. The value of $α_{3}$ is 3. 192

S. The value of $α_{4}$ is 4. 200

5. 381

6. 461

The correct option is: [2018]

P → 4; Q → 6; R → 2; S → 1
P → 1; Q → 4; R → 2; S → 3
P → 4; Q → 6; R → 5; S → 2
P → 4; Q → 2; R → 3; S → 1

1

2

3

4

(3)

$Given 6 boys M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6} and 5 girls G_{1}, G_{2}, G_{3}, G_{4}, G_{5}$

$(i) α_{1} \to Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls$

i.e., ${}^{6}C_{3} \times {}^{5}C_{2} = 20 \times 10 = 200$ $∴$ $α_{1} = 200$

$(ii) α_{2} \to Total number of ways of selecting at least 2 members having equal number of boys and girls$

i.e., ${}^{6}C_{1} {}^{5}C_{1} + {}^{6}C_{2} {}^{5}C_{2} + {}^{6}C_{3} {}^{5}C_{3} + {}^{6}C_{4} {}^{5}C_{4} + {}^{6}C_{5} {}^{5}C_{5}$

$= 30 + 150 + 200 + 75 + 6 = 461$ $∴$ $α_{2} = 461$

$(iii) α_{3} \to Total number of ways of selecting 5 members in which at least 2 of them are girls$

i.e., ${}^{5}C_{2} {}^{6}C_{3} + {}^{5}C_{3} {}^{6}C_{2} + {}^{5}C_{4} {}^{6}C_{1} + {}^{5}C_{5} {}^{6}C_{0}$

$= 200 + 150 + 30 + 1 = 381$ $∴$ $α_{3} = 381$

$(iv) α_{4} \to Total number of ways for selecting 4 members in which at least two girls such that M_{1} and G_{1} are not included together.$

$G_{1} is included$ $\to$ ${}^{4}C_{1} \cdot {}^{5}C_{2} + {}^{4}C_{2} \cdot {}^{5}C_{1} + {}^{4}C_{3} = 40 + 30 + 4 = 74$

$M_{1} is included$ $\to$ ${}^{4}C_{2} \cdot {}^{5}C_{1} + {}^{4}C_{3} = 30 + 4 = 34$

$G_{1} and M_{1} both are not included$ $= {}^{4}C_{4} + {}^{4}C_{3} \cdot {}^{5}C_{1} + {}^{4}C_{2} \cdot {}^{5}C_{2} = 1 + 20 + 60 = 81$

$∴$ $Total number = 74 + 34 + 81 = 189$ $∴$ $α_{4} = 189$

$Now, P \to 4; Q \to 6; R \to 5; S \to 2$

Q 15 :

Let $a_{n}$ denote 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 $c_{n}$ = the number of such $n$ -digit integers ending with digit 0. [2012]

Q. The value of $b_{6}$ is

7
8
9
11

1

2

3

4

(2)

$∵$ $a_{n} = number of all n -digit positive integers formed by the digits 0, 1 or both such that no consecutive digits are 0 .$

$and b_{n} = number of such n -digit integers ending with 1$

$and c_{n} = number of such n -digit integers ending with 0$

$Clearly, a_{n} = b_{n} + c_{n}$ $(∵ a_{n} can end with 0 or 1)$

$Also b_{n} = a_{n - 1}$

$and c_{n} = a_{n - 2}$ $[∵ if last digit is 0, second last has to be 1]$

$∴$ $a_{n} = a_{n - 1} + a_{n - 2}, n \geq 3$

$Also a_{1} = 1, a_{2} = 2$

$Now by this recurring formula, we get$

$a_{3} = a_{2} + a_{1} = 3$

$a_{4} = a_{3} + a_{2} = 3 + 2 = 5$

$a_{5} = a_{4} + a_{3} = 5 + 3 = 8$

$Also b_{6} = a_{5} = 8$

Q 16 :

Let $a_{n}$ denote 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 $c_{n}$ = the number of such $n$ -digit integers ending with digit 0. [2012]

Q. Which of the following is correct?

$a_{17} = a_{16} + a_{15}$
$c_{17} \neq c_{16} + c_{15}$
$b_{17} \neq b_{16} + c_{16}$
$a_{17} = c_{17} + b_{16}$

1

2

3

4

(1)

$By recurring formula, a_{17} = a_{16} + a_{15} is correct.$

$Also c_{17} \neq c_{16} + c_{15}$

$\Rightarrow a_{15} \neq a_{14} + a_{13}$ $(∵ c_{n} = a_{n - 2})$ $∴$ $Incorrect$

$Similarly, other parts are also incorrect.$

Topic Question Set

Q 1 :

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is [2016]

Q 2 :

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is [2012]

Q 3 :

A rectangle with sides of length ( $2 m - 1$ ) and ( $2 n - 1$ ) units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is [2005]

[IMAGE 27]

Q 4 :

Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals [2001]

Q 6 :

Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{y}{9 x} =$ [2017]

Q 8 :

Let $n \geq 2$ be an integer. Take $n$ distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of $n$ is [2014]

Q 12 :

Let $S = {1, 2, 3, \dots, 9}$ . For $k = 1, 2, \dots, 5$ , let $N_{k}$ be the number of subsets of $S$ , each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$ [2017]

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	LIST-I		LIST-II
P.	The value of $α_{1}$ is	1.	136
Q.	The value of $α_{2}$ is	2.	189
R.	The value of $α_{3}$ is	3.	192
S.	The value of $α_{4}$ is	4.	200
		5.	381
		6.	461

Topic Question Set

Q 1 : A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is [2016]

Q 2 : The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is [2012]

Q 3 : A rectangle with sides of length (2m−1) and (2n−1) units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is [2005] [IMAGE 27]

Q 4 : Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn+1-Tn=21, then n equals [2001]

Q 5 : A group of 9 students, s1,s2,…,s9 is to be divided to form three teams X, Y, and Z of sizes 2, 3, and 4, respectively. Suppose that s1 cannot be selected for the team X, and s2 cannot be selected for the team Y. Then the number of ways to form such teams is ________. [2024]

Q 6 : Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, y9x= [2017]

Q 8 : Let n≥2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is [2014]

Q 9 : Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in S, but 0210222 is NOT in S. Then the number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x, is equal to ______. [2025]

Q 12 : Let S={1,2,3,…,9}. For k=1,2,…,5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1+N2+N3+N4+N5= [2017]

Q 13 : For non-negative integers s and r, let (sr)={s!r!(s-r)!if r≤s,0if r>s. For positive integers m and n, let g(m,n)=∑p=0m+nf(m,n,p)(n+pp) where for any non-negative integer p, f(m,n,p)=∑i=0p(mi)(n+ip)(p+np-i). Then which of the following statements is/are TRUE? [2020]

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Q 1 :

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is [2016]

Q 2 :

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is [2012]

Q 3 :

A rectangle with sides of length ( $2 m - 1$ ) and ( $2 n - 1$ ) units is divided into squares of unit length by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is [2005]

[IMAGE 27]

Q 4 :

Let $T_{n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n + 1} - T_{n} = 21$ , then $n$ equals [2001]

Q 6 :

Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{y}{9 x} =$ [2017]

Q 8 :

Let $n \geq 2$ be an integer. Take $n$ distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of $n$ is [2014]

Q 12 :

Let $S = {1, 2, 3, \dots, 9}$ . For $k = 1, 2, \dots, 5$ , let $N_{k}$ be the number of subsets of $S$ , each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$ [2017]