Question

$16$ players ${P_1},\;{P_2},\;...,\;{P_{16}}$ take part in a tennis tournament. Lower suffix player is better than any higher suffix player. These players are to be divided into $4$ groups each comprising $4$ players and the best from each group is selected for the semi-finals.
Number of ways in which they can be divided into equal groups if the players ${P_1},\;{P_2},\;{P_3},\;{P_4}$ are in different groups, is?
(A) $\dfrac{{(11)!}}{{36}}$
(B) $\dfrac{{(11)!}}{{72}}$
(C) $\dfrac{{(11)!}}{{108}}$
(D) $\dfrac{{(11)!}}{{216}}$

Answer 1

We use the concept of factorials to figure out the exact number of ways we can divide the given players. This is because we require the concept of permutations and combinations. Since they have already given us the total number of players needed in each group we just need to remember that all the players have to be placed in some group.

Complete step by step solution:
Let us see the given details;
Total number of players considered is $= 16$ players
Total number of groups that must be made from the given players $= 4$ groups
And the total number of players within each of the $4$ groups are also $= 4$ players
They have also given us another detail that the first four players (${P_1},\;{P_2},\;{P_3},\;{P_4}$) are definitely going to be placed in four different groups. This is a key detail because this means that one player has already occupied one out of the four total available places in each of the four groups.
In simple words, there is one player in each group, let it be
${P_1}$ – Placed in first group
${P_2}$ - Placed in the second group
${P_3}$ - Placed in the third group
${P_4}$ - Placed in the fourth group
Now we just need to find where the other players will be placed.
So this means that the total number of players left to be placed $= 16 - 4 = 12$ players.
This implies that we can only place the remaining players in $12!$ ways.
But we know that every player must also be placed in a group and there are four groups. Within these four groups, three placements each are available so the total ways in which the groups can be formed is in $3!\; \times 3!\, \times 3!\; \times 3!$.
Therefore the ways in which equal groups can be formed from the remaining players are demonstrated as follows:
$\Rightarrow$ Total number of ways $= \dfrac{{12!}}{{3!\; \times 3!\, \times 3!\; \times 3!}}$
Simplifying we get:
$\Rightarrow$ Total number of ways $= \dfrac{{12 \times 11!}}{{3!\; \times 3!\, \times 3!\; \times 3!}}$
$\Rightarrow$ Total number of ways $= \dfrac{{11!}}{{3!\; \times 3!\, \times 3}}$
So the final answer is:
$\Rightarrow$ Total number of ways $= \dfrac{{11!}}{{108}}$
So it is evident that the options (A), (B) and (D) containing values $\dfrac{{(11)!}}{{36}}$, $\dfrac{{(11)!}}{{72}}$, $\dfrac{{(11)!}}{{216}}$ respectively are not the right answers because after solving the given problem, the resultant value was $\dfrac{{(11)!}}{{108}}$.
Therefore the correct answer is option (C) $\dfrac{{(11)!}}{{108}}$.

Note:
In the above question we were able to use permutations to find the total number of ways we could divide the given players, but keep in mind that when we perform permutations, we must not include the same player twice within one group. Sometimes we might consider the same player multiple times but always remember the total number of ways we can place a given number of people is equal to the factorial of number of people, for example: $3$ people can only be placed in $3!$ ways.