Will decides to attend a basketball game with four friends. If the par

Answer is C.
2x4x3x2x1 = 48

Sent from my SM-G935F using Tapatalk

Will decides to attend a basketball game with four friends. If the party of five sits together in five consecutive seats, and Will must NOT sit in between two of his friends, how many ways can the five friends be arranged?

(A) 24
(B) 36
(C) 48
(D) 72
(E) 120

total ways ; 5!
total ways in which will can sit b/w ; 4!
5!-4!/2 ; 48
IMO C

Will decides to attend a basketball game with four friends. If the party of five sits together in five consecutive seats, and Will must NOT sit in between two of his friends, how many ways can the five friends be arranged?

(A) 24
(B) 36
(C) 48
(D) 72
(E) 120

"Will must NOT sit in between two of his friends" means that he must st on ether end o the ve people .e.
1) W _ _ _ _
or
2) _ _ _ _ W

Hence
Case 1) Possibilities at the 5 positions respectively are
\(1 * 4 * 3 * 2 * 1 = 24\)

Since for 2nd position their are four of his friends to sit and anyone can take that place. After that 3rd position has 3 people left to take the place and for 4th position 2 are left and finally last one takes the last place.

Case 2) Possibilities at the 5 positions respectively are
\(4 * 3 * 2 * 1 * 1 = 24\)

Will takes place the fifth position since positions 1, 2, 3 & 4 are not for him as per condition laid out in question.
As explained for case 1 similarly, positions with their respective possibilities are filled up.

Thus, \(24 + 24 = 48\)

Answer (C).

5 friends: W(ill), A, B, C, D are having two major arrangements as follows:

(1) W x x x x
No of ways can the five friends be arranged = 4*3*2*1 = 24

(2) x x x x W
No of ways can the five friends be arranged = 4*3*2*1 = 24

(1) + (2) = 48 ways
Answer is (C)

Imo C is the right answer. 48

(Total possible ways) - (will sit in between) = Will not sit in between
5!- (4!x3)= 120-(24x3) = 48

Will can sit either to the right of all his friends or to the left of all his friends. This means there there are two possibilities for will to sit with all his four friends.

The 4 friends can sit in 4! arrangements
4!=4*3*2*1=24.
So total possible arrangements =2*24=48.
Answer is therefore C imo.

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.