Let S = { 1 , 2 , 3 , … , 9 } . For k = 1 , 2 , … , 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]

Question

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]

Smart Achievers · Accepted Answer

(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.

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

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

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

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

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

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

Solution

Q.
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]

1 210

2 252

3 125

4 126

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