Counting Problem
\(\large \color{black}{\begin{align} & \normalsize \text{4 men and 3 women are to be seated in row so that } \hspace{.33em}\\~\\ & \normalsize \text{no two women sit together. } \hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which they can be seated. } \hspace{.33em}\\~\\ \end{align}}\)
ok so we lsit the ways to do this can you think of any
\(\large \color{black}{\begin{align} & a.)\ 4!\times ^{5}P_{3} \hspace{.33em}\\~\\ & b.)\ 4!\times ^{5}C_{3} \hspace{.33em}\\~\\ & c.)\ 4!\times ^{4}P_{3} \hspace{.33em}\\~\\ & d.)\ 4!\times ^{4}C_{3} \hspace{.33em}\\~\\ \end{align}}\)
@mathmate
i have found a link can any one explan me http://www.meritnation.com/ask-answer/question/m-men-and-n-women-are-to-be-seated-in-a-row-so-that-no-two-w/permutations-and-combinations/5901983
Yes, it does help!
So we first seat the 4 men, so there are 4! ways of seating the gents. |dw:1440446507784:dw|
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