Question

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
10​

Answer 1

Answer:

in 63 ways groups can be formed

Step-by-step explanation:

Total number of men($n_{1}$) = 7

number of men to be chosen for group($c_{1}$) = 5

possible number of ways for choosing men($P_{1}$) = $\frac{n_{1}! }{c_{1}! * (n_{1} - c_{1})!}$ (Binomial Th.)

                                                                        = $\frac{7!}{5! * (7-5)!} = \frac{7!}{5! * 2!}$

                                                         =  $\frac{1*2*3*4*5*6*7}{(1*2*3*4*5)*(1*2)} = \frac{6*7}{1*2} = 3*7 = 21$

Total number of women($n_{2}$) = 3

number of women to be chosen for group($c_{2}$) = 2

possible number of ways for choosing women($P_{2}$) = $\frac{n_{2}! }{c_{2}! * (n_{2} - c_{2})!}$ (Binomial Th.)

                                                                               = $\frac{3!}{2! * (3-2)!} = \frac{3!}{2! * 1!} = \frac{3!}{2!}$

                                                                              =$\frac{1*2*3}{1*2} = \frac{3}{1} = 3$

Therefore number of ways in which groups can be formed with 5 men and 2 women out of 7 men and 3 women = $P_{1} * P_{2} = 21 *3 = 63$

Therefore in 63 ways groups can be formed