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

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Will decides to attend a basketball game with four friends. If the par  [#permalink]

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01 Sep 2019, 22:47
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25% (medium)

Question Stats:

82% (01:05) correct 18% (01:43) wrong based on 39 sessions

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Competition Mode Question

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

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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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01 Sep 2019, 23:16
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Here's how I did:

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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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01 Sep 2019, 23:24
1
seats are of the form - - - - -
now Will can either sit in extreme left or extreme right
leaving 4 arrangements 4! and consider 2 ways (extreme left or extreme right) hence 2 *4!
= 2 * 4*3*2 = 48
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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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01 Sep 2019, 23:50
1
2x4x3x2x1 = 48

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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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02 Sep 2019, 01:33
1
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
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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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02 Sep 2019, 04:22
1
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, $$24 + 24 = 48$$

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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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02 Sep 2019, 05:29
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
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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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02 Sep 2019, 06:48
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
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Re: Will decides to attend a basketball game with four friends. If the par  [#permalink]

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02 Sep 2019, 12:15
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.

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Re: Will decides to attend a basketball game with four friends. If the par   [#permalink] 02 Sep 2019, 12:15
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