Question

The number of ways of arranging 6 players to throw the cricket ball so that the oldest player may not throw first is:
A. 120
B. 600
C. 720
D. 7156

Answer 1

Hint: To find the number of ways of arranging 6 players to throw the cricket ball so that the oldest player may not throw first. First we will find the possibilities of throwing the first ball so that the oldest player wont through first. Then we will arrange the remaining player in the form of n! where, $n!=n\left( n-1 \right)\left( n-2 \right)...........1$. At last we will multiply $5\times n!$ to get the total number of ways.

Complete step-by-step solution -
It is given in the question that there are 6 players to throw the cricket ball. So the oldest player may not throw the ball first.
Let us consider ${{P}_{1}},{{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}}\And {{P}_{6}}$ are the 6 players from which ${{P}_{1}}$ is the oldest player.
As it is mentioned in the question that the oldest player should not throw the first, As ${{P}_{1}}$ is the oldest player then the possibilities of throwing the first ball are $6-1=5$ . So the possibilities of throwing the first ball is 5.
As the one of the players is already thrown, then the possibility of throwing a second ball is n i.e. 5.
As the two players have already thrown the ball, then the possibility of throwing the third ball is $\left( n-1 \right)$ i.e. 4.
As the three players have already thrown the ball, then the possibility of throwing fourth balls is $\left( n-2 \right)$ i.e. 3.
As the four players have already thrown the ball, then the probability of throwing the fifth ball is $\left( n-3 \right)$ i.e. 2. And so on.
We can also say that the next 5 throws will be 5!.
We know that, $n!=n\left( n-1 \right)\left( n-2 \right)...........1$
Therefore, required number of ways $=5\times 5!$
$\begin{align}
  & =5\times 5\times 4\times 3\times 2\times 1 \\
 & =600\text{ ways} \\
\end{align}$
Hence, option B. is the correct answer.

Note: Students can also solve this question in another way.
The total number arrangements of 6 player be $6!$
Where, $6!=6\times 5\times 4\times 3\times 2\times 1$
$=720$ Ways
If the oldest player through the first ball, then the arrangements of remaining 5 players be $5!.$
Where, $5!=5\times 4\times 3\times 2\times 1$
$=120$ Ways
Therefore, the number of ways of arrangements so that oldest player may not through ball first will be
$720-120=600$ Ways.