Question

A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then?
$(A)$ m= n= 78
$\left( B \right)$ n = m-8
$\left( C \right)$ m + n= 68
$\left( D \right)$ m = n = 68

Answer 1

Hint: Apply the formula of ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

In this question it is given to us that we have a total 13 members i.e. 8 males and 5 females and with the help of that we have to form a committee of 11 members.
Where m= Number of ways the committee is formed at least 6 males.
So in this case we’ll have at least 6 males i.e. it could be either 6 males or 7 males or 8 males and in that case we’ll have either 5 females or 4 females or 3 females and hence we have,
m= $\left( {{}^8{C_6} \times {}^5{C_5}} \right) + \left( {{}^8{C_7} \times {}^5{C_4}} \right) + \left( {{}^8{C_8} \times {}^5{C_3}} \right)$
m= $(28 \times 1) + (8 \times 5) + (1 \times 10)$
And hence we have,
m = 78
and n= number of ways the committee is formed with at least 3 females
Now similarly we have at least 3 female it means that it could be either 3 females or 4 females or 5 females and in that case we’ll have either 8 males or 7 males or 6 males and hence we have,
n${\text{ = (}}{}^5{C_3} \times {}^8{C_8}) + {\text{(}}{}^5{C_4} \times {}^8{C_7}) + {\text{(}}{}^5{C_5} \times {}^8{C_6})$
n$= (10 \times 1) + (5 \times 8) + (1 \times 28)$
and hence on solving, we have
n = 78
and hence the value of m and n is 78.
Therefore option A i.e. m=n=78 is the correct answer.

Note: In this type of question we have to make proper selection and after making proper selection we just need to apply the formula of ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and hence on putting the value over in this formula and on doing the simplification, we’ll have our answer.