The number of ways in which 15 identical gold | vedclass

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(B) Let $x_1, x_2, x_3$ be the number of coins received by the $3$ persons.We have the condition $x_1 + x_2 + x_3 = 15$ where $x_i \geq 3$ for $i = 1,

Vedclass · Accepted Answer

$28$

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(B) Let $x_1, x_2, x_3$ be the number of coins received by the $3$ persons.We have the condition $x_1 + x_2 + x_3 = 15$ where $x_i \geq 3$ for $i = 1, 2, 3$.Let $y_i = x_i - 3$,then $y_i \geq 0$.Substituting $x_i = y_i + 3$ into the equation:$(y_1 + 3) + (y_2 + 3) + (y_3 + 3) = 15$$y_1 + y_2 + y_3 + 9 = 15$$y_1 + y_2 + y_3 = 6$The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 6$ and $r = 3$.Number of ways $= \binom{6+3-1}{3-1} = \binom{8}{2} = \frac{8 	imes 7}{2 	imes 1} = 28$.

The number of ways in which $15$ identical gold coins can be distributed among $3$ persons such that each one gets at least $3$ gold coins is

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