Question

In how many ways 10 boys and 10 girls can be arranged such that they are alternate

Answer 1

Hi,

Answer:

11! * 10! ways arrangement can be done

Step-by-step explanation:

Let us consider the boys to occupy 10 places at first,  

So, this can be done in 10! ways.

It is given that the boys and girls are to be arranged alternately.

After the arrangement of the boys we have 9 places in between each of the boys and 1 place before the first boy and 1 place after the last boy.  

Therefore, there are 10 girls to be placed in 11 places i.e., ¹¹P₁₀ = 11! / (11-10)! = 11! ways

∴ No. of ways boys and girls can be arranged alternately = 11! * 10! Ways  

Hope this helps!!!!!

Answer 2

Answer: 144,850,083,840,000 ways

Step-by-step explanation:

Let us first made an arrangement of seating of girls as shown

_G_G_G_G_G_G_G_G_G_G_

here we can easily seen that 10 girls are seated and the ways by which they can sit are 10! ways.

Now we have 11 vacant seats ,there we have to sit 10 boys ;

n= 11

r= 10

number of ways by which those boys can be arranged

$^{11} P_{10} = \frac{11!}{(11 - 10)!} = 11!$

Hence ,10! 11! many ways 10 boys and 10 girls can be arranged such that they are alternate.

Hope it helps you.