Question

You are getting a line up ready for a school kickball game. You have 6 girls and 6 boys. The rules state each child must kick the same number of times and alternate girl-boy or boy-girl. How many ways can a line-up be made for one round of kicking?

Answer 1

Concepts and reason

Factorial of a number: The factorial of a non-negative integer n (say) is defined by the product of all non-negative integers less than or equal to n.

Permutation: In mathematical statistics, the term permutation refers to the fact of arranging the units of a set in different possible order. However, permutation signifies the count that how many times the arrangement of the units of the set can be altered.

Combination: Stands for selection of things. It stands for picking up a subset out of a set. Combination doesn’t include the concept of order of the things.

Fundamentals

The formula for factorial of a non-negative integer is given by,

$n! = n(n - 1)(n - 2)...1$

Permutation: A permutation is a choice of r things out of n things when the choice matters. The formula for permutation is given by,

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

Combination: A combination is a choice of r things out of n things when the choice does not matter. The formula for combination is given by,

$_n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}$

The objective of the problem is to determine the number of ways in which the boys and girls can be arranged so that each one must kick alternatively that is ‘girl-boy or boy-girl’ and they should same number of times. Also, the number of boys is 6 and girls are 6.

The number of ways in which the boys and girls can be arranged so that each child must kick the same number of times and alternate girl-boy or boy-girl is obtained as shown below:

Let B denote a boy and G denote a girl. The arrangement stated in the information can be obtained in the following two ways.

$\left\{ {BG{\rm{ }}BG{\rm{ }}BG{\rm{ }}BG{\rm{ }}BG{\rm{ }}BG\;} \right\}$or $\left\{ {GB{\rm{ }}GB{\rm{ }}GB{\rm{ }}GB\;GB{\rm{ }}GB} \right\}$.

In both cases, the boys can be permuted in$6!$ ways and the girls can also be permuted into $6!$ ways. Also there are two possible ways of arrangement.

The number of ways is,

$\begin{array}{c}\\{\rm{Number of ways}} = 2 \times 6! \times 6!\\\\ = 2 \times 720 \times 720\\\\ = 1,036,800\\\end{array}$

Ans:

Thus, the number of ways in which the boys and girls can be arranged so that each child must kick the same number of times and alternate girl-boy or boy-girl is 1,036,800