In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

Hindi

6.Permutation and Combination

easy

In how many ways can $5$ girls and $3$ boys be seated in a row so that no two boys are together?

$14400$

Solution

Let us first seat the $5$ girls. This can be done in $5 !$ ways. For each such arrangement, the three boys can be seated only at the cross marked places.

$\times G \times G \times G \times G \times G \times$

There are $6$ cross marked places and the three boys can be seated in $^{6} P _{3}$ ways. Hence, by multiplication principle, the total number of ways

${ = 5!{ \times ^6}{P_3} = 5! \times \frac{{6!}}{{3!}}}$

${ = 4 \times 5 \times 2 \times 3 \times 4 \times 5 \times 6 = 14400}$

Standard 11

Mathematics

If $2 \times {}^n{C_5} = 9\,\, \times \,\,{}^{n – 2}{C_5}$, then the value of $n$ will be

easy

$^{n – 1}{C_r} = ({k^2} – 3)\,.{\,^n}{C_{r + 1}}$ if $k \in $

hard

(IIT-2004)

To fill $12$ vacancies there are $25$ candidates of which five are from scheduled caste. If $3$ of the vacancies are reserved for scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made

medium

The number of ways to give away $25$ apples to $4$ boys, each boy receiving at least $4$ apples, are

normal

All possible numbers are formed using the digits $1, 1, 2, 2, 2, 2, 3, 4, 4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is

hard

(JEE MAIN-2019)