Question

There are 6 batsmen, 7 bowlers and 2 wickets keepers. Find the number of ways of forming a team of 11 players having at least 4 batsmen, at least 5 bowlers and at least 1 wicket keeper.
(a) $567$
(b) $525$
(c) $462$
(d) $777$

Answer 1

Hint: Before we proceed with the problem, it is important to know the definitions of combinations because we will be using this concept. We will have 3 cases to solve this question. In the first case, we will be having 4 batsmen, 5 bowlers, and 2 wicket-keepers. In the second case, 4 batsmen, 6 bowlers, and 1 wicket-keeper, and in the third case, 5 batsmen and 5 bowlers and 1 wicketkeeper.

Formula Used: ${}^n{C_r}$

Complete step by step Solution:
Given that batsmen = 6; bowlers = 7; wickets keepers = 2 and Team Size =11 players
Case 1: We take 4 batsmen and 5 bowlers and 2 wicket-keepers.
So, the number of ways of forming a team, in this case, is given by:
$ \Rightarrow {}^6{C_4} \times {}^7{C_5} \times {}^2{C_2}$
$ \Rightarrow (6 \times \dfrac{5}{2}) \times (7 \times \dfrac{6}{2}) \times 1 = 315$
Case 2: We take 4 batsmen and 6 bowlers and 1 wickets keeper
So, the number of ways of forming a team, in this case, is given by:
$ \Rightarrow {}^6{C_4} \times {}^7{C_6} \times {}^2{C_1}$
$ \Rightarrow (6 \times \dfrac{5}{2}) \times (7) \times 2 = 210$
Case 3: We take 5 batsmen and 5 bowlers and 1 wickets keeper

So, the number of ways of forming a team, in this case, is given by:
$ \Rightarrow {}^6{C_5} \times {}^7{C_5} \times {}^2{C_1}$
$ \Rightarrow (6) \times (7 \times \dfrac{6}{2}) \times 2 = 252$
Now we will add all the ways of all the cases.
So total number of ways $ = 315 + 210 + 252$
                                                        $ = 777$
Therefore, the total number of ways of forming a team of 11 players having at least 4 batsmen, at least 5 bowlers and at least 1 wicketkeeper is 777.

Hence, the correct option is d.

Note: In these types of questions, we should always use combination formulas. Do not forget to add the result of all the 3 cases. Combinations cannot be confused with permutations. In permutations, the order of the selected items is essential.