In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected ?

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Solution

Total men = 6

Total women = 4

Total persons in committee = 5

(where at least are women has to be selected)

This can be done in

$^{4} C_{1} \times^{6} C_{4} +^{4} C_{2} \times^{6} C_{3} +^{4} C_{3} \times^{6} C_{2} +^{4} C_{4} \times^{6} C_{1}$

$(^{n} C_{r} = \frac{n!}{r! (n - r)!}) (^{n} C_{r} = 1,^{n} C_{1} = n)$

$= (\frac{4 \times 6!}{4! \times 2!}) + (\frac{4!}{2! 2!} \times \frac{6!}{3! 3!}) + (\frac{4!}{3! 1!} \times \frac{6!}{2! 4!}) + (1 \times 6)$

$= (\frac{4 \times 6 \times 5}{2}) + (\frac{4 \times 3}{2} \times \frac{6 \times 5 \times 4}{3 \times 2}) + (\frac{4 \times 6 \times 5}{2}) + (6)$

$= (60) + (120) + 60 + 6$

= 246 ways

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