Question

In how many ways a team of 11 must be selected from 5 men and 11 women such that the team comprises of not more than three men?

Answer 1

Answer:

2256

Step-by-step explanation:

In how many ways a team of 11 must be selected from 5 men and 11 women such that the team comprises of not more than three men?

Men can be  3    ,  2   , 1  or 0

Women then be  8  , 9  , 10  & 11

⁵C₃*¹¹C₈  + ⁵C₂*¹¹C₉  + ⁵C₁*¹¹C₁₀ + ⁵C₀*¹¹C₁₁

= 10 * 165  + 10*55 + 5*11  + 1

= 1650  + 550  + 55 + 1

= 2256

Answer 2

Answer:

Total ways to comprises team of 11 by taking not more than 3 men = 2256

Step-by-step explanation:

According to the question

The selection can be made like

3M 8W---11 member

2M 9W---11 member

1M 10W---11 member

0M 11W---11 member

Now to choose 3 men from 5 can be given by

$^{5} C_3$

8 women from 11

$^{11} C_{8}$

So select the team of 11

$^{5} C_3 \times ^{11} C_{8} + ^{5} C_2 \times ^{11} C_{9} + ^{5} C_1 \times ^{11} C_{10} +^{5} C_0 \times^{11} C_{11} \\ \\ \frac{5!}{3! \times 2!} \times \frac{11!}{8! \times 3!} + \frac{5!}{2! \times 3!} \times \frac{11!}{9! \times 2!} + \frac{5!}{4!} \times \frac{11!}{10!} + 1 \\ \\ \frac{5 \times 4 \times 3!}{3! \times 2} \times \frac{11 \times 10 \times 9 \times 8!}{8! \times 3 \times 2} + \frac{5 \times 4\times 3!}{3! \times 2} \times \frac{11 \times 10 \times 9!}{9! \times 2} + 5 \times 11 + 1 \\ \\ 10 \times 165 + 10 \times 55 + 55 + 1 \\ \\ 1650 + 550 + 56 \\ \\ 2256$