In how many ways a committee of 5 members can be selected from 6 men a

1.	In how many ways a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?1). 1202). 2203). 2004). 320
Answer» Overall we have to select 5 members in all, among these 5, there should be 3 MEN and 2 ladies. Now we have to select 3 men for the committee from the group of 6 men i.e. ⁶C₃ = $(\frac{{6!}}{{3!\LEFT( {6 - 3} \right)!}})$ $(= \frac{{6!}}{{3!.3!}} = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 6}} = 20)$ Similarly, we have to select 2 ladies for the committee from the group of 5 ladies i.e. ⁵C₂ = $(\frac{{5!}}{{2!\left( {5 - 2} \right)!}} = \frac{{5!}}{{2!3!}} = \frac{{120}}{{2 \times 6}} = 10)$ Total NUMBER of POSSIBLE selections= 20 × 10 = 200

In how many ways a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?1). 1202). 2203). 2004). 320

Answer»

Overall we have to select 5 members in all, among these 5, there should be 3 MEN and 2 ladies.

Now we have to select 3 men for the committee from the group of 6 men i.e. ⁶C₃ = $(\frac{{6!}}{{3!\LEFT( {6 - 3} \right)!}})$

$(= \frac{{6!}}{{3!.3!}} = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 6}} = 20)$

Similarly, we have to select 2 ladies for the committee from the group of 5 ladies i.e. ⁵C₂ = $(\frac{{5!}}{{2!\left( {5 - 2} \right)!}} = \frac{{5!}}{{2!3!}} = \frac{{120}}{{2 \times 6}} = 10)$

Total NUMBER of POSSIBLE selections= 20 × 10 = 200

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