Solveeit Logo
Login

Question

Question: The total number of ways in which \({n^2}\) number of identical balls can be put in ‘n’ numbered box...

The total number of ways in which n2{n^2} number of identical balls can be put in ‘n’ numbered boxes (1, 2, 3, 4,.......,n) such that the ithi^{th} box contains at least ‘i’ number of balls is.
A. n2Cn−1^{{n^2}}{C_{n - 1}}
B. n2−1Cn−1^{{n^2} - 1}{C_{n - 1}}
C. n2+n−12Cn−1^{\dfrac{{{n^2} + n - 1}}{2}}{C_{n - 1}}
D. None of these

Explanation

Solution

Here in this question we will first write the given condition in sequence form determining the total number of ways in which n2{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)=nCrC(n,r) = {}^n{C_r}
C(n,r)=n!r!(n−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:-
⇒x1+x2+x3+......+xn=n2\Rightarrow {x_1} + {x_2} + {x_3} + ...... + {x_n} = {n^2} Such that, xi≥i,i∈1{x_i} \ge i,i \in 1to n
Now let xi−i=yi{x_i} - i = {y_i} or we can write this as xi=yi+i{x_i} = {y_i} + i
Since xi≥i,i∈1{x_i} \ge i,i \in 1then yi≥i{y_i} \ge i
Now after putting xi=yi+i{x_i} = {y_i} + iin the original equation it becomes as,
⇒(y1+1)+(y2+2)+........+(yn+n)=n2,yi≥0\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.
⇒y1+y2+........+yn=n2−(n+(n−1)+......+2+1)\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. Sn=n2[2a+(n−1)d]{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=Tn−Tn−1d = {T_n} - {T_{n - 1}}
⇒(n+(n−1)+......+2+1\Rightarrow (n + (n - 1) + ...... + 2 + 1 Here a=n,d=n−1−n=−1a = n,d = n - 1 - n = - 1
Now applying these values in Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}[2a + (n - 1)d]
⇒Sn=n2[2n+(n−1)(−1)]\Rightarrow {S_n} = \dfrac{n}{2}[2n + (n - 1)( - 1)]
⇒Sn=n2[2n−n+1]\Rightarrow {S_n} = \dfrac{n}{2}[2n - n + 1]
⇒Sn=n2[n+1]\Rightarrow {S_n} = \dfrac{n}{2}[n + 1]
Now we will put this sum value in main equation.
⇒y1+y2+........+yn=n2−n2[n+1]\Rightarrow {y_1} + {y_2} + ........ + {y_n} = {n^2} - \dfrac{n}{2}[n + 1]
Now we will solve right hand side equation.
⇒y1+y2+........+yn=n2−(n2+n)2\Rightarrow {y_1} + {y_2} + ........ + {y_n} = {n^2} - \dfrac{{({n^2} + n)}}{2}
⇒y1+y2+........+yn=2n2−(n2+n)2=(n2−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−1Cr−1^{n + r - 1}{C_{r - 1}} where n is the sum and r is the number of variables.
Therefore total possibilities= (n2−n)2+n−1Cn−1\dfrac{{({n^2} - n)}}{2}{ + ^{n - 1}}{C_{n - 1}}
⇒(n2−n)2+n−1Cn−1{ \Rightarrow ^{\dfrac{{({n^2} - n)}}{2} + n - 1}}{C_{n - 1}}
⇒n2−n+2n−22Cn−1=n2+n−22Cn−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 n2{n^2} number of identical balls is n2+n−22Cn−1^{\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.