In this question, n girls are to be seated alternatively between m boys.
There are m+1 spaces in which girls can be arranged.
The number of ways of arranging n girls is P(m+1,n) = \(\frac{(m+1)!}{(m-n+1)!}\) ways.
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n,r) = n!/(n-r)!
Therefore, permutation of n different objects in m+1 places is
P(m+1,n) = \(\frac{(m+1)!}{(m+n-1)!}\)
= \(\frac{(m+1)!}{(m-n+1)!}\)
The arrangement of m boys can be done in P(m, m) ways.
Formula:
Number of permutations of n distinct objects among r different places, where repetition is not allowed, is
P(n, r) = n!/(n-r)!
Therefore, a permutation of m different objects in m places is
P(m, m) = \(\frac{m!}{(m-m)!}\) = \(\frac{m!}{0!}\) = m!
Therefore the total number of arrangements is \(\frac{(m+1)!}{(m-n+1)!}\times m!\)