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Question: The number of solutions of \[{x_1} + {x_2} + {x_3} = 51\]( \[\left( {{x_1},{x_2},{x_3}} \right)\] be...

The number of solutions of x1+x2+x3=51{x_1} + {x_2} + {x_3} = 51( (x1,x2,x3)\left( {{x_1},{x_2},{x_3}} \right) being odd natural numbers) is

A.300300

B.325325

C.330330

D.350350

Explanation

Solution

Let the odd natural number be in the forms of 2a+1,2b+1,2c+12a + 1,2b + 1,2c + 1, where a, b, c are the natural numbers. Hence, we can change (x1,x2,x3)\left( {{x_1},{x_2},{x_3}} \right) variables with the above formed variables. And hence we can apply the concept of permutation and combination as when r things are to be distributed among n particular persons then the formula applied is n+r−1Cr−1^{n + r - 1}{C_{r - 1}}. Hence, finally expand the value of n+r−1Cr−1^{n + r - 1}{C_{r - 1}}by using the formula of nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}.

Complete step-by-step answer:
As per the given, the equation is x1+x2+x3=51{x_1} + {x_2} + {x_3} = 51( (x1,x2,x3)\left( {{x_1},{x_2},{x_3}} \right) being odd natural numbers)
Let the odd numbers be 2a+1,2b+1,2c+12a + 1,2b + 1,2c + 1, respectively.
Substituting the values of all the odd numbers in x1+x2+x3=51{x_1} + {x_2} + {x_3} = 51 equation as
⇒2a+1+2b+1+2c+1=51\Rightarrow 2a + 1 + 2b + 1 + 2c + 1 = 51
On simplifying the equation,
⇒2a+2b+2c=48\Rightarrow 2a + 2b + 2c = 48
Now on dividing the equation by 2 and on simplifying the above equation we get,
a+b+c=482=24a + b + c = \dfrac{{48}}{2} = 24
Now, as we know have to distribute 2424 things among 33 particulars quantities,
So now applying the concept that when r things is to be distributed among n particular persons then the formula applied is n+r−1Cr−1^{n + r - 1}{C_{r - 1}}. So here n=24,r=3n = 24,r = 3
On substituting the value, in the above equation, we get,
⇒24+3−1C3−1=26C2{ \Rightarrow ^{24 + 3 - 1}}{C_{3 - 1}}{ = ^{26}}{C_2}
So, on expanding the above expression using the formula of nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}},
⇒26C2=26!2!(24)!{ \Rightarrow ^{26}}{C_2} = \dfrac{{26!}}{{2!\left( {24} \right)!}}
Hence, on simplifying we get,
{ \Rightarrow ^{26}}{C_2}$$$$ = \dfrac{{26 \times 25 \times 24!}}{{2!\left( {24} \right)!}}
On cancellation of common terms we get,

{ \Rightarrow ^{26}}{C_2}$$$$ = \dfrac{{26 \times 25}}{2}
On further simplification we get,
{ \Rightarrow ^{26}}{C_2}$$$$ = 13 \times 25
On multiplication we get,
{ \Rightarrow ^{26}}{C_2}$$$$ = 325
Hence the number of solutions of x1+x2+x3=51{x_1} + {x_2} + {x_3} = 51 where (x1,x2,x3)\left( {{x_1},{x_2},{x_3}} \right) being odd natural numbers, are 325325.
Hence, option (B) is our correct answer.

Note: Number of ways in which n identical things can be divided into r groups, if blank groups are allowed (here groups are numbered, i.e., distinct)
==Number of ways in which n identical things can be distributed among r persons, each one of them can receive 0,1,20,1,2 or more items
=n+r−1Cr−1{ = ^{n + r - 1}}{C_{r - 1}}
Hence, remember the expansion of formula and calculate the equation properly.