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Q23E

Expert-verifiedFound in: Page 414

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**How many ways are there for eight men and five women to stand in a line so that no two women stand next to each other? (Hint: First position the men and then consider possible positions for the women.)**

The total number of possible arrangements is \(609,638,400\).

Number of men \( = 8\) and number of women \( = 5\).

**The word "permutation" refers to the act or process of changing the linear order of an ordered set.**

**Formula:**

\(_n{P_r} = \frac{{n!}}{{(n - r)!}}\)

First, consider the position of men.

Find the possible ways to arrange men in a row.

\(\begin{array}{l}P(8,8) = \frac{{8!}}{{(8 - 8)!}}\\P(8,8) = \frac{{8!}}{{0!}}\\P(8,8) = 40,320\end{array}\)

It is given that no two women stand next to each other.

The situation becomes:

\(O\;M\;O\;M\;O{\rm{ }}M\;O\;M\;O\;M\;O{\rm{ }}M\;O\;M\;O\;M\;O\)

There are \(9\) places for women. We can arrange \(5\) women in these 9 places.

Now, find the ways to place women:

\(\begin{array}{l}P(9,5) = \frac{{9!}}{{(9 - 5)!}}\\P(9,5) = \frac{{9!}}{{4!}}\\P(9,5) = 15,120\end{array}\)

Find the total number of possible arrangements:

Number of possible arrangements \(\; = 15,120 \times 40,320\)

Number of possible arrangements \( = 609,638,400\)

Hence, the total number of possible arrangements is \(609,638,400\).

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