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men and women are to be seated in a row so that no two women sit together. If then the number of ways in which they can be seated is:
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a
m!(m+1)!(m−w+1)!
b
mCm−w(m−w)!
c
m+wCm(m−w)!
d
none of these
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Correct option is A
We first arrange the m men. This can be done in m! ways. After m men have taken their seats, the women must choose w seats out of (m + 1) seats marked with X below.XMXMXMX…XMX 1st 2nd 3rd m th (m+1) th They can choose w seats in m+1Cw ways and arrange w women in w! ways.Thus, the required number of arrangements ism! m+1Cw(w!)=m!(m+1)!w!w!(m+1−w)!=m!(m+1)!(m+1−w)!Similar Questions
The number of ways in which we can get at least 6 successive tails when a coin is tossed 10 times in a row, is
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