There are ten boys B_{1}, B_{2}, ldots, B_{10 | vedclass

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(B) Total number of boys $n(B) = 10$ and total number of girls $n(G) = 5$. The total number of ways to form a group of $3$ boys and $3$ girls without

Vedclass · Accepted Answer

$1120$

Vedclass · Answer

(B) Total number of boys $n(B) = 10$ and total number of girls $n(G) = 5$. The total number of ways to form a group of $3$ boys and $3$ girls without any restrictions is: $= {}^{10}C_{3} 	imes {}^{5}C_{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} 	imes \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = 120 	imes 10 = 1200$. Now,calculate the number of ways where both $B_{1}$ and $B_{2}$ are members of the group. If $B_{1}$ and $B_{2}$ are already selected,we need to choose $1$ more boy from the remaining $8$ boys and $3$ girls from the $5$ girls: $= {}^{8}C_{1} 	imes {}^{5}C_{3} = 8 	imes 10 = 80$. The number of ways where $B_{1}$ and $B_{2}$ are not both in the same group is the total ways minus the restricted ways: $= 1200 - 80 = 1120$.

There are ten boys $B_{1}, B_{2}, \ldots, B_{10}$ and five girls $G_{1}, G_{2}, \ldots, G_{5}$ in a class. The number of ways of forming a group consisting of three boys and three girls,such that $B_{1}$ and $B_{2}$ are not both members of the same group,is

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