Question

If 3n different things can be equally distributed among 3 persons in k ways then calculate the number of ways to divide the 3n things in 3 equal groups.
A) \[kx3!\]
B) $\dfrac{k}{{3!}}$
C) ${(3!)^k}$
D) None of these

Answer 1

Hint: We solve this by using the concept of combination. A selection (group) of number of things (objects) taking some or all at a time is called combination. The total number of combinations of n distinct things taking $r(1 \leqslant r \leqslant n)$ at a time is denoted by $^n{C_r}$ .

Complete step by step solution:
It is given that 3n different things can be equally distributed among 3 persons in k ways, which means there are 3n things and we are distributed among 3 persons and let us consider r to be the equal amount of thing which is distributed to each person such that 1st person is getting r thing, 2nd person is getting r thing and 3rd person is also getting r thing and all these things can be distributed in k ways.
Then by Combination formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
But here we have 3n as total number of things, so we replace all the places of n as 3n.
So, ${}^{3n}{C_r} = \dfrac{{3n!}}{{r!\left( {3n - r} \right)!}} = k$
$\therefore k = \dfrac{{3n!}}{{r!\left( {3n - r} \right)!}}$ …… (1)
Now we have to calculate the number of ways to divide the 3n things in 3 equal groups which means we have to find the number of ways to divide 3n things in 3 equal groups and in every group number of things equal say r.
Now since we are dividing the 3n things equally in 3 groups we will have the value obtained by the combination formula multiplied by$3!$ because we give things to first group we have 3 choices , then one choice is reduced, then we give things to second group which has two choices and again one choice is reduced so the last group has no choice, so number of ways to distribute things among three groups is \[3 \times 2 \times 1 = 3!\]
So, $3! \times {}^{3n}{C_r} = k$
Dividing both sides by \[3!\].
$\therefore \dfrac{k}{{3!}} = $ number of ways

Hence, the option (B) is correct.

Note:
Students many times forget that things can be distributed inside the three groups also so they miss onto the step where we multiply with \[3!\]. Students many times make mistake when opening the factorial terms, so always remember that \[n! = n(n - 1)! = n(n - 1)(n - 2)!..... = n(n - 1)(n - 2)(n - 3).....3.2.1\].
$^n{C_r}$ can also be denoted as ${}^n{C_r}$ or $\left( {\mathop {}\limits_r^n } \right)$
Properties of combination:
(1) ${}^n{C_r} = {}^n{C_{n - r}}$
 (2) ${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$ $1 \leqslant r \leqslant n$
 (3) If ${}^n{C_x} = {}^n{C_y}$, then either $x=y$ or $x + y= n$