Question

There are 8boys of different heights. in how many ways can 8boys be arranged so that the tallest and shortest boys are not together

Answer 1

Given:

The given information is there are 8 boys of different height.

To Find:

We have to find the number of ways in which these 8 boys of different heights can be arranged so that the tallest and shortest are not together.

Solution:

Let us consider a,b,c,d,e,f,g,h are the 8 boys, where 'a' is the tallest boy and 'h' is the shortest boy.

Since 'a' is the tallest boy who is standing at the first position and 'h' is the shortest boy, who is standing at the last position, so, we have 6 boys to arrange.

Now,

1) Number of ways to arrange the remaining 6 boys $=6!$

Then,

2) Number of ways to arrange the two boys among the 6 boys = $\[{}_{2}^{7}\text{C}\]$

Also, the boys (a,h) can exchange their position, so,

3) Number of ways to arrange the two boys among the 6 boys $=2!$

Thus,

Number of ways to arrange the 8 boys such that the tall and the short boy are not together $\[=\left( 2! \right)\times \left( 6! \right)\times {}_{2}^{7}\text{C}\]$ $= 30240$

Hence, the required number of ways to arrange the 8 boys such that the tall and the short boy are not together is 30240.

Answer 2

Given : There are 8boys of different heights.

To Find : in how many ways can 8boys be arranged so that the tallest and shortest boys are not together

Solution:

8 boys can be arranged in 8! Ways

8!

Assume that tallest and shortest boys are together consider them as 1 who can be arranged in 2 ways

now 1 this group and 6 remaining

Hence total = 7

Can be arranged in 7!.2 Ways

Ways  8 boys can  be arranged so that the tallest and shortest boys are not together  = 8! -  7!.2

= 7! ( 8 - 2)

=  7!  .6

= 30240

30240 Ways  8 boys be arranged so that the tallest and shortest boys are not together

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