Question

In a group of 13 cricket players four are bowlers find out in how many ways can they form a cricket team of 11 players in which at least 2 bowlers are in included
A.55
B.72
C.78
D.None of these

Answer 1

Hint- In this question we use the theory of permutation and combination. So, before solving this question we need to first recall the basics of this chapter. For example, if we need to select 2 players out of four. Then in this case, this can be done in${}^{\text{4}}{{\text{C}}_2}$ =6 ways.

Complete step-by-step answer:
Now, as given in the question,
13 cricket players and 4 bowlers are available and we need to select 11 players in such a way at least two bowlers.
Remaining players = 13-4=09
As we know, ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
Total ways = select two bowlers + select three bowlers + select four bowlers
= ${}^9{C_9} \times {}^4{C_2} + {}^9{C_8} \times {}^4{C_3} + {}^9{C_7} \times {}^4{C_4}$
= 6+36+36
= 78
Therefore, a total of 78 ways can form a cricket team of 11 players in which at least 2 bowlers are included.
Thus, option (C) is the correct answer.

Note- We need to remember this formula for selecting r things out of n.
                                                       ${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n.
for example, $3!{\text{ = 3}} \times {\text{2}} \times 1$