**Question:**

A man has 10 friend And wants to arrange them on chair taking 6 At a time.find the number of arrangement such that 1.two particular friend are always taken 2.four particular friends are never taken

1)= 8c4×6!

2)= 8c6×6!

Take 6 as a group and arrange remaining 5 + 1 as object so (6-1)! is __ans__

For the second half you can think in this way total arrangements are 10c6*6! Now since 4 are never together so arrangements for being together will be 10c6*4!*3! Therefore the answer for never together should be 10c6*6!-10c6*4!*3!

According to me first select 6 people from 10 in 10c2 ways now there are 6 remaining positions two are fixed so what you can do is arrange rest 4 in 4! __Ways__

10c2*2!*8p3

1.select 4 from 8 as 2 are already selected in 8c4 ways n now arrange the 6 friends in 6! way

Ans=8c4×6!

2. 4 frnz are never taken so only remaining 6 frnz are taken n we also need 6 frnz only so need of selection only we hv to arrange which we can do in 6!ways

Ans=6!

ans. of ques 1. is 8c6&

ques.2 is 1

1 10c6×4!

2 10c6 (6!-3!)

Answer

1- 2*8P4

2-6!

10P6 is total no. Of possible arrangements. For 1st,2×8P4n for2nd,10P6-4