A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done, if the committee should have 4 professors and 1 research associate or all 3 trainees and 2 professors?
A. 22
B. 13
C. 24
D. 52
E. None of these
Answer: Option E
Solution(By Examveda Team)
Required number of ways$$ = \left( {{}^4{C_4} \times {}^6{C_1}} \right) + $$ $$\left( {{}^3{C_3} \times {}^4{C_{2}}} \right)$$
$$\eqalign{ & = \left( {1 \times 6} \right) + \left( {1 \times \frac{{4 \times 3}}{2}} \right) \cr & = \left( {6 + 6} \right) \cr & = 12 \cr} $$
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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!
yes answer is not 14 it is 12
How come fourteen... it’s should be twelve I think
= (4C4X6C1) + (3C3X4C2)
= (1X6) + (1X4*3/2)
= 6+6 = 12