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(B) To get a sum of $7$ with $5$ dice,each die must show at least $1$. Let the outcomes be $x_1, x_2, x_3, x_4, x_5$ where $x_i \ge 1$. The sum is $x_

Vedclass · Accepted Answer

$15$

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(B) To get a sum of $7$ with $5$ dice,each die must show at least $1$. Let the outcomes be $x_1, x_2, x_3, x_4, x_5$ where $x_i \ge 1$. The sum is $x_1 + x_2 + x_3 + x_4 + x_5 = 7$. Since each $x_i \ge 1$,we can write $x_i = 1 + y_i$ where $y_i \ge 0$. Substituting this,we get $(1+y_1) + (1+y_2) + (1+y_3) + (1+y_4) + (1+y_5) = 7$,which simplifies to $y_1 + y_2 + y_3 + y_4 + y_5 = 2$. The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$ where $n=2$ and $r=5$. This is $\binom{2+5-1}{5-1} = \binom{6}{4} = \binom{6}{2} = \frac{6 	imes 5}{2} = 15$. Alternatively,the cases are: $(i)$ Four $1$s and one $3$: $\frac{5!}{4!1!} = 5$ ways. (ii) Three $1$s and two $2$s: $\frac{5!}{3!2!} = 10$ ways. Total ways = $5 + 10 = 15$.

If $5$ dice are rolled simultaneously,then the number of ways of getting a total of $7$ on their faces is:

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