Question

In how many ways can a cricket team of 11 players selected out of 16 players if two particular players are to be included and one particular player is to be rejected?
a.715
b.615
c.915
d.515

Answer 1

Hint: As one player is not to be chosen so total players become 15 players and as we know 2 are particularly to be selected. So, players to be selected become $\left( 11-2 \right)\ =9$ from $\left( 15-2 \right)\ =13$ players.

Complete step-by-step answer:

In the question we are asked to find in how many ways a cricket team of 11 players which are to be selected out of 16 players given the condition that two players are to be included and one particular player should be rejected.
This question is related to chapter permutation and combination, where topic combination is to be used.
Before proceeding we will first learn what combination is. In mathematics, a combination is a solution of items from a collection, such that (unlike permutations) the order of selection does not matter. For example, given three fruits, say an apple, orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear, an apple and an orange, a pear and an orange. More formally a K- combination of set S is a subset of k-distinct elements of S. If the set has ${{n}^{-}}$ elements, the number K- combinations is equal to binomial coefficient.

n \\\ k \\\ \end{matrix} \right)\ =\ \dfrac{n\left( n-1 \right)......\left( n-k+1 \right)}{k\left( k-1 \right).............1}$$ Which can be written using factorials as $$\dfrac{n!}{k!\left( n-k \right)!}$$ wherever $$k\le n$$ and which is zero when $$k>n$$. The set of all k- combinations of a set S is often denoted by $$\left( \begin{matrix} s \\\ k \\\ \end{matrix} \right)$$. Now, in the question it was asked to select 11 players out of 16 players with some given conditions in which two particular players are to be selected and one has to be rejected. So, by the conditions as we know one has to be rejected and two players has to be taken so total players becomes 13, and as two are particularly chosen the number of players to be chosen is 9. So, total number of ways are $$^{13}{{\text{C}}_{\text{9}}}$$, which can be calculated as $$\dfrac{13!}{9!\times 4!}$$ which on calculation we get 715. So, the correct option is ‘a’. Note: Students should read the questions properly and also know how to apply the conditions, to ensure that the solution does not go wrong.