The number of ways in which 21 identical appl | vedclass

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(D) Let the number of apples given to the three children be $x_1, x_2, x_3$ respectively.We have $x_1 + x_2 + x_3 = 21$,where $x_i \ge 2$ for $i = 1,

Vedclass · Accepted Answer

$136$

Vedclass · Answer

(D) Let the number of apples given to the three children be $x_1, x_2, x_3$ respectively.We have $x_1 + x_2 + x_3 = 21$,where $x_i \ge 2$ for $i = 1, 2, 3$.Let $y_i = x_i - 2$,so $y_i \ge 0$.Substituting $x_i = y_i + 2$,we get $(y_1 + 2) + (y_2 + 2) + (y_3 + 2) = 21$.$y_1 + y_2 + y_3 + 6 = 21$,which simplifies to $y_1 + y_2 + y_3 = 15$.The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 15$ and $r = 3$.Number of ways = $\binom{15+3-1}{3-1} = \binom{17}{2}$.$\binom{17}{2} = \frac{17 	imes 16}{2 	imes 1} = 17 	imes 8 = 136$.

The number of ways in which $21$ identical apples can be distributed among three children such that each child gets at least $2$ apples,is

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