Question

The total number of ways in which ${n^2}$ number of identical balls can be put in ‘n’ numbered boxes (1, 2, 3, 4,.......,n) such that the $i^{th}$ box contains at least ‘i’ number of balls is.
A. $^{{n^2}}{C_{n - 1}}$
B. $^{{n^2} - 1}{C_{n - 1}}$
C. $^{\dfrac{{{n^2} + n - 1}}{2}}{C_{n - 1}}$
D. None of these

Answer 1

Hint: Here in this question we will first write the given condition in sequence form determining the total number of ways in which ${n^2}$ balls will get arranged then we will apply the following permutation and combination formula. These are mentioned below: -
Combination: -Number of combination of ‘n’ things has taken ‘r’ at a time then combination formula is given by: -
$C(n,r) = {}^n{C_r}$
$C(n,r) = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step by step solution:
So if we write the given problem in equation form then it can be written as:-
$ \Rightarrow {x_1} + {x_2} + {x_3} + ...... + {x_n} = {n^2}$ Such that, ${x_i} \ge i,i \in 1$to n
Now let ${x_i} - i = {y_i}$ or we can write this as ${x_i} = {y_i} + i$
Since ${x_i} \ge i,i \in 1$then ${y_i} \ge i$
Now after putting ${x_i} = {y_i} + i$in the original equation it becomes as,
$ \Rightarrow ({y_1} + 1) + ({y_2} + 2) + ........ + ({y_n} + n) = {n^2},{y_i} \ge 0$
Now we will arrange y terms together while n terms on the right hand side of the equal to sign.
$ \Rightarrow {y_1} + {y_2} + ........ + {y_n} = {n^2} - (n + (n - 1) + ...... + 2 + 1)$
Now we will solve right hand side of the equation by applying arithmetic formula i.e. ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$ where a= first term of the sequence and d=common difference which is given by $d = {T_n} - {T_{n - 1}}$
$ \Rightarrow (n + (n - 1) + ...... + 2 + 1$ Here $a = n,d = n - 1 - n = - 1$
Now applying these values in ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
$ \Rightarrow {S_n} = \dfrac{n}{2}[2n + (n - 1)( - 1)]$
$ \Rightarrow {S_n} = \dfrac{n}{2}[2n - n + 1]$
$ \Rightarrow {S_n} = \dfrac{n}{2}[n + 1]$
Now we will put this sum value in main equation.
$ \Rightarrow {y_1} + {y_2} + ........ + {y_n} = {n^2} - \dfrac{n}{2}[n + 1]$
Now we will solve right hand side equation.
$ \Rightarrow {y_1} + {y_2} + ........ + {y_n} = {n^2} - \dfrac{{({n^2} + n)}}{2}$
$ \Rightarrow {y_1} + {y_2} + ........ + {y_n} = \dfrac{{2{n^2} - ({n^2} + n)}}{2} = \dfrac{{({n^2} - n)}}{2}$ (Taking L.C.M)
Now total possibilities is given by $^{n + r - 1}{C_{r - 1}}$ where n is the sum and r is the number of variables.
Therefore total possibilities= $\dfrac{{({n^2} - n)}}{2}{ + ^{n - 1}}{C_{n - 1}}$
${ \Rightarrow ^{\dfrac{{({n^2} - n)}}{2} + n - 1}}{C_{n - 1}}$
${ \Rightarrow ^{\dfrac{{{n^2} - n + 2n - 2}}{2}}}{C_{n - 1}}{ = ^{\dfrac{{{n^2} + n - 2}}{2}}}{C_{n - 1}}$ (Taking L.C.M in the upper part of the formula)
Therefore the total number of ways in which ${n^2}$ number of identical balls is $^{\dfrac{{{n^2} + n - 2}}{2}}{C_{n - 1}}$ .

So, the correct answer is “Option C”.

Note: Students may likely make mistakes in applying combination properties so they should know the combination concept because without that these types of questions can create little difficulties.