Math, asked by tonydevathoti2000, 6 months ago

Jennifer has 10 boxes with her.In how many ways can she arrange them so that 4 of the boxes do not remain together.?

Answers

Answered by amitnrw
4

Given :  Jennifer has 10 boxes with her. She arrange them so that 4 of the boxes do not remain together

To find :  Number of ways she can arrange

Solution :

10 boxes  can be arranged   in 10!  ways

4 of the boxes do not remain together  

= Total ways -  4 of the boxes remain together  

10 boxes = 6 Boxes + 4 Boxes

4 of the boxes remain together  

 take 4 boxes as  1 box which can be arranged in 4!  ways

Total boxes = 6 + 1 =  7 boxes  

7 Boxes can be arranged in 7!  Ways

Total ways of arranging  4 of the boxes remain together    

7!.4!

4 of the boxes do not remain together  = Total ways -  4 of the boxes remain together  

=  10!  -7!.4!

= 7! ( 10 * 9 * 8 - 24)

= 7! ( 696)

= 35,07,840

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Answered by Vyomsingh
6

Given:

  • Jennifer has 10 boxes

________________________________

To Find:

  • Number of ways to arrange those boxes so that 4 of the boxes do not remain together.

________________________________

Solution:

The number of ways the 10 boxes can be arranged.➛

10! ways.

✏️If we subtract the possibility of the 4 boxes remaining together

such as:-

4 boxes = 1 unit ,

Remaining 6 boxes = 6 units .

Total = 7 units

7 units can be arranged in 7! ways

and the 4 boxes can be arranged among themselves in 4! ways.

Therefore, the answer is 10!-(7!*4!).

_____________________________

Answer:

10!-(7!*4!)

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