Question

A man wants to distribute 101 coins of 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. ${}^{103}{C}_2-3{\times }^{52}{C}_2$
• B. $\frac{{}^{103}{C}_2}{3}$
• C. $1275$
• D. $\frac{{}^{103}{C}_2}{6}$

Answer 1

Answer: (A), (C)

The correct options are
A ${}^{103}{C}_2-3{\times }^{52}{C}_2$
C $1275$

Let the amount received by the sons be Rs. x , Rs. y and Rs. z, respectively, then
x $\le$ y + z = 101 - x
i.e., 2x $\le$ 101 $\therefore$ x $\le$50, y $\le$50, z $\le$50
x + y + z = 101
The corresponding multinomial is $(1+x+x^2+....x^{50}{)}^3$
Hence total number of distributions 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 of ${\left(\frac{x^{51}-1}{x-1}\right)}^3$
$=$coefficient of $x^{101}$ in the expansion of $-{\left(x^{51}-1\right)}^3(1-x{)}^{-3}$

$=$coefficient of $x^{101}$ in the expansion of $-(x^{153}-1-3x^{102}+3x^{51})(1+3x{+}^4{C}_2{x}^2{+}^5{C}_3{x}^3+..{.}^{52}{C}_{50}{x}^{50}+...{+}^{103}{C}_{101}{x}^{101}+...)$

${=}^{103}{C}_{101}-{3.}^{52}{C}_{50}$
${=}^{103}{C}_2-{3.}^{52}{C}_2$

$=1275$