Question

How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other? (Hint: First position the women and then consider possible positions for the men.)

Short answer

The total number of possible arrangements is \(1,207,084,032,000\).

Step-by-step solution

Given data

Number of women \( = 10\) and number of men \( = 6\)

Concept of Permutation

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

Formula:

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

Calculation to find the position of men

First, consider the position of women.

Find the possible ways to arrange women in a row:

\(\begin{array}{l}P(10,10) = \frac{{10!}}{{(10 - 10)!}}\\P(10,10) = \frac{{10!}}{{0!}}\\P(10,10) = 3,628,800\end{array}\)

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

The situation becomes:

\(O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\;W\;O\)

Calculation to find the number of possible arrangement

There are \(11\) places for men. We can arrange \(6\) men in these \(11\) places.

Now, find the ways to place men:

\(\begin{array}{l}P(11,6) = \frac{{11!}}{{(11 - 6)!}}\\P(11,6) = \frac{{11!}}{{5!}}\\P(11,6) = 332,640\end{array}\)

Find the total number of possible arrangements:

Number of possible arrangements \( = 332,640 \times 3,628,800\)

Number of possible arrangements \( = 1,207,084,032,000\)

Hence, the total number of possible arrangements is \(1,207,084,032,000\).