Math, asked by hemanthsivasai19, 2 days ago

Five students A,B,C ,D and E are getting bored of their regular study. They go to the playground and sit in a straight line. Answer the following; i)Find the number of ways of sitting if A and D can sit together. ii)Find the total number of arrangements if B is sitting in the middle or if E is sitting in the extreme left.​

Answers

Answered by aditimaji2011
2

Answer:

Solution

verified

Verified by Toppr

Ram → R Shyam → S

No. of ways of arranging in which R & S together =4!×2!=48

No. of ways of arranging in which R & S not together = Total ways to arrange − No. of ways of arranging in which R & S together.

=5!−4!×2!

=120−48=72.

Answered by ajajit9217
8

Answer:

There are 48 ways of sitting if B is sitting in the middle or if E is sitting in the extreme left.​

Step-by-step explanation:

(i) We need to find the number of ways provided A and D sit together.

Let A and D be consider as a single unit say X

Now we need to arrange B , C , E , X

They can be arranged in 4! ways.

X itself can be arranged in 2! ways.

Therefore, the number of ways of sitting if A and D can sit together = 4! * 2!

= 48 ways.

Therefore, there are 48 ways of sitting if A and D can sit together.

(ii) Number of arrangements if B is sitting in the middle or E is on the extreme left.

= When B is in the middle + When E is on the extreme left.

No. of ways of sitting when B is in the middle ( _ _ B _ _ )

= 4 * 3 * 1 * 2 * 1 = 24 ways

No. of ways of sitting when E is on the extreme left ( E _ _ _ _ )

= 1 * 4 * 3 * 2 * 1 = 24 ways

Therefore, the total arrangements = 24 + 24 = 48 ways.

Therefore, there are 48 ways of sitting if B is sitting in the middle or if E is sitting in the extreme left.​

                                                                             

Similar questions