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Find the number of ways in which m boys and n girls may be arranged in a row so that no two of the girls are together; it is given that m > n.

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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 \(\frac{(m+1)!}{(m-n+1)!}\times m!\)

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