How many different ways can 5 girls and 5 boys?

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Hint: Proceed the solution of this question, using the fundamental theorem which is the number of ways of arranging n unlike objects in a line is n! and for the case of circular arrangement use (n - 1)!.

Complete step-by-step answer:

(a) Here we are considering boys and girls as unlike objects hence we can use the fundamental theorem to find the number of ways of making the line in which boys and girls alternate.

Fundamental theorem- Number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’).

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

5 boys can be arranged in a line by 5! ways.

Here the boys and girls are alternating.

Let the boys be B1 , B2 , B3 , B4 , B5

And they can be arranged by 5! Ways and leaves spaces shown by cross

⇒B1 × B2 × B3 × B4 × B5 ×

We are left with 5 places marked by a cross where five girls can be arranged in 5! ways. Thus by fundamental theorem it will be 5! × 5! = \[{\left( {5!} \right)^2}\]

Now the position (1) could be

We are left with 5 places marked by a cross where five girls can-be arranged in 5! ways. Thus by fundamental theorem it will be 5! × 5! = \[{\left( {5!} \right)^2}\]

Hence the total number of ways will be \[{\left( {5!} \right)^2} + {\left( {5!} \right)^2} = 2 \times {\left( {5!} \right)^2} = 240\]

In other words, we can say if we start the line with the boy in the 1st position as in (1) there are \[{\left( {5!} \right)^2}\] ways & if we start with girl in 1st position as in (2) then also there are \[{\left( {5!} \right)^2}\] ways.

Thus the total number of ways will be

⇒ \[{\left( {5!} \right)^2} + {\left( {5!} \right)^2} = 2 \times {\left( {5!} \right)^2} = 28800\]

(b) We know that the total number of circular arrangements of n persons is = (n - 1)!

First arrange the girls in a circle. (In this case it doesn't really matter whether the girls are seated first or last, because the number of girls is the same as the number of boys.)

The number of ways of doing this is (5-1)! =4! Ways, since this is a cyclic permutation.

After the girls are seated, then since boys and girls have to alternate, there will be 5 places, one place each for boys in between the girls. These 5 places can be filled by 5 boys in 5! Ways.

So, the required number. of ways = 5!$ \times $4!

= 5$ \times $4$ \times $3$ \times $2$ \times $1$ \times $4$ \times $3$ \times $2$ \times $1

Therefore, there are 2,880 ways of forming a circle where 5 boys and 5 girls sit alternatingly.

Note: In this particular question we should know that the permutation in a row has a beginning and an end, but there is nothing like beginning or end in circular permutation. Thus, in circular permutation, we consider one object is fixed and the remaining objects are arranged in (n - 1)! ways (as in the case of arrangement in a row).

And conceptually we can say that one circular arrangement corresponds to an unique row (linear) arrangement.

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(Last Updated On: March 12, 2019)

Problem Statement: EE Board October 1997

How many different ways can 5 boys and 5 girls form a circle with boys and girls alternate?

Problem Answer:

There are 2,880 ways can 5 boys and 5 girls form a circle with boys and girls alternate.

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First arrange the girls in a circle. (In this case it doesn't really matter whether the girls are seated first or last, because the number of girls is the same as the number of boys.) The number of ways of doing this is 4! = 24, since this is a cyclic permutation. After the girls are seated, there are 5 spaces in-between where the boys can be inserted. There are 5! = 120 ways of doing this. Therefore there are 24*120= 2,880 ways of forming a circle where 5 boys and 5 girls sit alternatingly.