Question

Two groups of players consist of 6 and 8 players. In how many ways can a team of 11 players be selected from these two groups if at least 4 players are to be included from the first group?
(a) 334
(b) 344
(c) 120
(d) 168

Answer 1

To solve this problem we will make three cases i.e. first when 4 players are taken from the first team and 7 players from the second one. Second case when 5 players are chosen from first team and 6 from second team. Third case when all 6 players are chosen from first and rest from second team. These all cases can be easily found out using $^{n}{{C}_{r}}$ where $^{n}{{C}_{r}}$ is equal to $\dfrac{n!}{r!\left( n-r \right)!}$. After finding out the total number of ways of selecting players from each team we will add all the number of cases to get the required answer.

Complete step-by-step answer:
We are given two groups of players first which consists of 6 players and second consists 8 players.
Now we have to find the total number of ways in which a team of 11 players can be formed which contains at least 4 players from group 1.
So to solve this we will form 3 cases as follows,
Case 1: when 4 players are selected from group 1 and 7 are selected from group 2.
So we get,
Number of ways of selecting 4 players from 6 players from group 1 = $^{6}{{C}_{4}}$
Number of ways of selecting 7 players from 8 players from group 2 = $^{8}{{C}_{7}}$
So total number of cases for this case we get as =
$\begin{aligned} & ^{6}{{C}_{4}}{{\times }^{8}}{{C}_{7}} \\\ & =\dfrac{6!}{4!\left( 6-4 \right)!}\times \dfrac{8!}{7!\left( 8-7 \right)!} \\\ & =\dfrac{6!}{4!\left( 2 \right)!}\times \dfrac{8!}{7!\left( 1 \right)!} \\\ \end{aligned}$
$\begin{aligned} & =15\times 8 \\\ & =120 \\\ \end{aligned}$
Case 2: when 5 players are selected from group 1 and 6 are selected from group 2.
Number of ways of selecting 5 players from 6 players from group 1 = $^{6}{{C}_{5}}$
Number of ways of selecting 6 players from 8 players from group 2 = $^{8}{{C}_{6}}$
So total number of cases for this case we get as = $^{6}{{C}_{5}}{{\times }^{8}}{{C}_{6}}=6\times 28=168$
$\begin{aligned} & ^{6}{{C}_{5}}{{\times }^{8}}{{C}_{6}} \\\ & =\dfrac{6!}{4!\left( 6-5 \right)!}\times \dfrac{8!}{7!\left( 8-6 \right)!} \\\ & =\dfrac{6!}{4!\left( 1 \right)!}\times \dfrac{8!}{7!\left( 2 \right)!} \\\ & =6\times 28 \\\ & =168 \\\ \end{aligned}$

Case 3: when 6 players are selected from group 1 and 5 are selected from group 2.
Number of ways of selecting 6 players from 6 players from group 1 = $^{6}{{C}_{6}}$
Number of ways of selecting 5 players from 8 players from group 2 = $^{8}{{C}_{5}}$
So total number of cases for this case we get as = $^{6}{{C}_{6}}{{\times }^{8}}{{C}_{5}}=1\times 56=56$
$\begin{aligned} & ^{6}{{C}_{6}}{{\times }^{8}}{{C}_{5}} \\\ & =\dfrac{6!}{4!\left( 6-5 \right)!}\times \dfrac{8!}{7!\left( 8-5 \right)!} \\\ & =\dfrac{6!}{6!\left( 0 \right)!}\times \dfrac{8!}{7!\left( 3 \right)!} \\\ & =1\times 56 \\\ & =56 \\\ \end{aligned}$
Note that value of 0! Is 1.
So we get,
Total number of ways of selecting team of 11 players which consists at least 4 players from group 1 as,
= 120 + 168 + 56 = 344

So, the correct answer is “Option (b)”.

Note: You need to first observe the question carefully to solve this and then only you will be able to understand how the cases are to be formed to solve it easily otherwise you may try to solve it directly without cases and get stuck up or end up with the wrong answer. And some students may also just add the number of ways of selecting players instead of multiplying so be careful about that too.