From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least three men are in the committee. In how many ways can it be done? a.624 b.702 c.756 d.812

Answer:

[tex]\qquad\qquad\qquad\boxed{ \bf{ \:(c) \: \: 756 \: \: }} \\ \\ [/tex]

Step-by-step explanation:

Given that, From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least three men are in the committee.

The following cases arises :

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf Number \: of \: men & \bf Number \: of \: women \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 3 & \sf 2 \\ \\ \sf 4 & \sf 1 \\ \\ \sf 5 & \sf 0 \end{array}} \\ \end{gathered} \\ \\ [/tex]

So, Number of ways are

[tex]\sf \: = \ ^7C_3 \times \ ^6C_2 + \ ^7C_4 \times \ ^6C_1 + \ ^7C_5 \times \ ^6C_0 \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{7!}{3!4!} \times \dfrac{6!}{2!4!} + \dfrac{7!}{4!3!} \times \dfrac{6!}{1!5!} + \dfrac{7!}{5!2!} \times 1 \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{7.6.5.4!}{3!4!} \times \dfrac{6.5.4!}{2!4!} + \dfrac{7.6.5.4!}{4!3!} \times \dfrac{6.5!}{5!} + \dfrac{7.6.5!}{5!2!} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{7.6.5}{3.2.1} \times \dfrac{6.5}{2.1} + \dfrac{7.6.5}{3.2.1} \times 6 + \dfrac{7.6}{2.1} \\ \\ [/tex]

[tex]\sf \: = \:35 \times 15 + 35 \times 6 + 21 \\ \\ [/tex]

[tex]\sf \: = \:525 + 210 + 21 \\ \\ [/tex]

[tex]\sf \: = \:756 \\ \\ [/tex]

Hence,

[tex]\bf\implies \: Number \: of \: ways \: = \: 756 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \: \ ^nC_r \: = \: \dfrac{n!}{r! \: (n - r)!} \\ \\ [/tex]

[tex]\sf \: \ ^nC_0 \: = \: 1 \\ \\ [/tex]

Answer:

756

Step-by-step explanation:

here 5 members are needed to be selected. out of that atleast 3 should be men

which means 3 m or 2 w, 4m or 1 w , 5m or 0w can be selected

hence total no of ways =7c3*6c2+7c4*6c1+7c5*6c0

solving the above equation we will get 756 ways