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…

Question

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=

MARKS Content Team · Accepted Answer

Correct answer: (D) 126. N_1+N_2+N_3+N_4+N_5= Total ways – {when no odd} Total ways =^9C_5 Number of ways when no odd, is zero (∵ only available even are 2, 4, 6, 8) ∴ ^9C_5-zero=126

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

Solution

Relevant Questions

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=

Solution

Relevant Questions

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