In a hotel,four rooms are available. Six pers | vedclass

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(C) Let the number of persons in the four rooms be $n_1, n_2, n_3, n_4$. Since each room has at least $1$ and at most $2$ persons,and the total number

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(C) Let the number of persons in the four rooms be $n_1, n_2, n_3, n_4$. Since each room has at least $1$ and at most $2$ persons,and the total number of persons is $6$,we have $n_1 + n_2 + n_3 + n_4 = 6$ where $1 \le n_i \le 2$.This implies two rooms must have $2$ persons and two rooms must have $1$ person (since $2+2+1+1 = 6$).The number of ways to distribute $6$ distinct persons into these groups is given by the multinomial coefficient $\frac{6!}{2!2!1!1!}$.Since the rooms are distinct,we must multiply by the number of ways to assign these group sizes to the $4$ rooms,which is $\frac{4!}{2!2!} = 6$.Thus,the total number of ways is $\frac{6!}{2!2!1!1!} 	imes 6 = \frac{720}{4} 	imes 6 = 180 	imes 6 = 1080$.

In a hotel,four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is . . . . . . . .

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