A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
A. 11340
B. 12750
C. 40
D. 320
Answer: Option A
Solution(By Examveda Team)
There 10 questions in part A out of which 8 question can be chosen as = 10C8 Similarly, 5 questions can be chosen from 10 questions of Part B as = 10C5 Hence, total number of ways, $$\eqalign{ & { = ^{10}}{{\text{C}}_8}{ \times ^{10}}{{\text{C}}_5} \cr & = \frac{{10!}}{{2! \times 8!}} \times \frac{{10!}}{{5! \times 5}} \cr & = \left\{ {10 \times \frac{9}{2}} \right\} \times \left\{ {\frac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}}} \right\} \cr & = 11340 \cr} $$Join The Discussion
Comments ( 2 )
Related Questions on Permutation and Combination
A. 3! 4! 8! 4!
B. 3! 8!
C. 4! 4!
D. 8! 4! 4!
A. 7560,60,1680
B. 7890,120,650
C. 7650,200,4444
D. None of these
A. 8 × 9!
B. 8 × 8!
C. 7 × 9!
D. 9 × 8!
Why is multiply not adding at the end ?
Wrong answer...
11340 is the right one