Question

A man wants to distribute 101 coins a rupee each, among his 3 sons with the condition that no one receives more money than the combined total of other two. The number of ways of doing this is :-
A. ${}^{100}{C_2} - 2 \times {}^{52}{C_2}$
B. $\dfrac{{{}^{100}{C_2}}}{3}$
C. 1275
D. \[\dfrac{{{}^{103}{C_2}}}{6}\]

Answer 1

Hint: The sum of the amount with three sons will be equal to 101. We will further apply the condition given that no one receives more money than the combined total of the other two.

Complete step-by-step answer:
Given, the amount is $Rs.101$
Now, let the amount of three sons be $x$, $y$ and $z$ respectively.
Given condition is $x + y + z = 101$ and $x \leqslant y + z$
Since,
$
  x + y + z = 101 \\
  x = 101 - (y + z) \\
$
Now, $x \leqslant y + z$
$
   \Rightarrow x \leqslant 101 - x \\
   \Rightarrow 2x \leqslant 101 \\
   \Rightarrow x \leqslant 50 \\
 $
Similarly, $y \leqslant 50$ and $z \leqslant 50$
And, $x + y + z = 101$
We will apply the concept of multinomial to find out the number of ways of the above-mentioned condition.
The corresponding Multinomial is ${(1 + x + {x^2} + ...{x^{50}})^3}$
Hence, total number of distribution is equivalent to coefficient of ${x^{101}}$ in the expansion of ${(1 + x + {x^2} + ...{x^{50}})^3}$
Coefficient of ${x^{101}}$ in the expansion ${(1 + x + {x^2} + ...{x^{50}})^3}$
The expansion ${(1 + x + {x^2} + ...{x^{50}})^3}$ can be simplified with the help of concept of sum of Geometric progression i.e., Sum $ = \dfrac{{a(1 - {r^n})}}{{1 - r}}$ where, a is the first term and r is the common ratio.
Applying,
 ${(1 + x + {x^2} + ...{x^{50}})^3} = {(\dfrac{{{x^{51}} - 1}}{{x - 1}})^3}$
The coefficient of ${x^{101}}$ in the expansion of ${(\dfrac{{{x^{51}} - 1}}{{x - 1}})^3}$
The coefficient of ${x^{101}}$ in the expansion of ${({x^{51}} - 1)^3}{(1 - x)^{ - 3}}$
The coefficient of ${x^{101}}$ in the expansion of$({x^{153}} - 1 - 3{x^{102}} + 3{x^{51}})(1 + 3x + {}^4{C_2}{x^2} + {}^5{C_3}{x^3} + ... + ...)$
The coefficient of ${x^{101}}$ is ${}^{103}{C_{101}} - 3{}^{52}{C_{50}}$
Hence, the coefficient of ${x^{101}}$ is 1275.

So, the correct answer is “Option C”.

Note: In the multinomial theorem, the sum is ${n_{1,}}{n_2},...,{n_k}$ taken over such that $n{}_1 + {n_2} + {n_3} + ... + {n_k} = n$. The multinomial theorem gives us a sum of multinomial coefficients multiplied by variables. In other words, it represents an expanded series where each term in it has its own associated multinomial coefficient.