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Question: The school cricket team has 11 players. The school wants a photograph of these players along with th...

The school cricket team has 11 players. The school wants a photograph of these players along with the principal and two vice-principals of the school for the school magazine. Seven chairs are arranged in a row for photography. Three chairs are reserved for the principal and two vice principals. Four players will occupy the remaining chairs and seven players will stand behind the chairs. The question is “in how many different ways can the eleven players take positions for the photograph”?

Explanation

Solution

In the above problem, basically we have to arrange one principal and two vice-principals and 11 players in such a way that three middle chairs will only occupy by the principal and vice-principal and the remaining 4 chairs out of 7 are occupied by players and also the remaining 7 players who are standing behind the chairs. The answer is, first of all, we are going to arrange two vice principals and one principal in the three middle chairs. Then we are going to select 4 players from 11 players by the combinatorial method and then do the permutations of these 4 players. Multiply the arrangement of principals, vice-principals with the arrangement of players arrangement. Now, the remaining 7 players which are standing behind the chairs are arranged using permutation. Multiply this permutation also with the previous arrangements.

Complete step-by-step solution:
First of all, we are arranging two vice principals and principals and 4 students on the 7 chairs. We are representing the chairs by spaces in the below:

In the above diagram, the middle three chairs are occupied by the principal and two vice principals so arranging these three people is done as follows:
3!3! ……… (1)
Now, 4 chairs are vacant and we have to fill 4 players in it so first of all, we are going to select 4 players out of 11 players using a combinatorial method.
11C4{}^{11}{{C}_{4}}
And we have to arrange these 4 students also so the arrangement of 4 players is 4!4! and multiplication of this arrangement by the above selection we get,
11C4.4!{}^{11}{{C}_{4}}.4!………. (2)
Now, we are representing the remaining 7 players which are standing behind the chairs as follows:

In the above diagram, we have represented the players standing behind the chairs by green dashes so arranging 7 players is equal to:
7!7! …….. (3)
Multiplying (1), (2) ,(3) will give us the total number of possible photographs that could be taken:
3!(11C44!)(7!) =3.2.1(11!4!7!.4!)(7!) \begin{aligned} & 3!\left( {}^{11}{{C}_{4}}4! \right)\left( 7! \right) \\\ & =3.2.1\left( \dfrac{11!}{4!7!}.4! \right)\left( 7! \right) \\\ \end{aligned}
In the above expression, 4!and7!4!and7! will be cancelled out from the numerator and denominator and we get,
6(11!) =239500800 \begin{aligned} & 6\left( 11! \right) \\\ & =239500800 \\\ \end{aligned}
Hence, 239500800 photographs are possible.

Note: The mistake that could happen in the above problem is that you might arrange the two vice principals because you think that these two names are similar but the catch is that names are similar but two vice principals are completely two different people so we have to arrange these two vice principals also.