A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 women. In how many different ways can it be done ?
A. 364
B. 728
C. 931
D. 1001
E. None of these
Answer: Option C
Solution(By Examveda Team)
Required number of ways$$ = \left( {{}^6{C_1} \times {}^8{C_3}} \right) + \left( {{}^6{C_2} \times {}^8{C_2}} \right)$$ $$ + \left( {{}^6{C_3} \times {}^8{C_1}} \right)$$ $$ + \left( {{}^6{C_4} \times {}^8{C_0}} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\} + $$ $$\left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{8 \times 7}}{{2 \times 1}}} \right)$$ $$ + \left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 8} \right)$$ $$ + \left( {{}^6{C_2} \times 1} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\}$$ $$ +\, 420\, + $$ $$\left( {\frac{{6 \times 5 \times 4}}{6} \times 8} \right)$$ $$ + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times 1} \right)$$
$$ = \left( {336 + 420 + 160 + 15} \right)$$
$$ = 931$$
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!
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