Question

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

A.\[300\]

B.\[325\]

C.\[330\]

D.\[350\]

Answer 1

Hint: Let the odd natural number be in the forms of \[2a + 1,2b + 1,2c + 1\], where a, b, c are the natural numbers. Hence, we can change \[\left( {{x_1},{x_2},{x_3}} \right)\] 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 \[^{n + r - 1}{C_{r - 1}}\]. Hence, finally expand the value of \[^{n + r - 1}{C_{r - 1}}\]by using the formula of \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].

Complete step-by-step answer:
As per the given, the equation is \[{x_1} + {x_2} + {x_3} = 51\]( \[\left( {{x_1},{x_2},{x_3}} \right)\] being odd natural numbers)
Let the odd numbers be \[2a + 1,2b + 1,2c + 1\], respectively.
Substituting the values of all the odd numbers in \[{x_1} + {x_2} + {x_3} = 51\] equation as
\[ \Rightarrow 2a + 1 + 2b + 1 + 2c + 1 = 51\]
On simplifying the equation,
\[ \Rightarrow 2a + 2b + 2c = 48\]
Now on dividing the equation by 2 and on simplifying the above equation we get,
\[a + b + c = \dfrac{{48}}{2} = 24\]
Now, as we know have to distribute \[24\] things among \[3\] particulars quantities,
So now applying the concept that when r things is to be distributed among n particular persons then the formula applied is \[^{n + r - 1}{C_{r - 1}}\]. So here \[n = 24,r = 3\]
On substituting the value, in the above equation, we get,
\[{ \Rightarrow ^{24 + 3 - 1}}{C_{3 - 1}}{ = ^{26}}{C_2}\]
So, on expanding the above expression using the formula of \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\],
\[{ \Rightarrow ^{26}}{C_2} = \dfrac{{26!}}{{2!\left( {24} \right)!}}\]
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 \[{x_1} + {x_2} + {x_3} = 51\] where \[\left( {{x_1},{x_2},{x_3}} \right)\] being odd natural numbers, are \[325\].
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 \[0,1,2\] or more items
\[{ = ^{n + r - 1}}{C_{r - 1}}\]
Hence, remember the expansion of formula and calculate the equation properly.