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=
A
126
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
125
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
210
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
252
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A126 Required number of subsets =5C1×4C4+5C2×4C3+5C3×4C2+5C4×4C1+5C5×4C0 =5+40+60+20+1=126