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Seven children are playing musical chairs. When the music stops, in ho

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Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 21 Mar 2017, 04:53
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Seven children are playing musical chairs. When the music stops, in how many ways can five of the seven children be arranged in a circle of five chairs?

A. 21
B. 42
C. 360
D. 504
E. 2520

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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 21 Mar 2017, 05:02
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7C5 x (N-1)! = 21 x 4! = 504
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 23 Mar 2017, 17:25
Here is my approach:

Because we're dealing with a circle, (5-1)! = 4*3*2*1 = 24 different seating arrangements

Next, (7!)/(5!)(2!) = 21 different ways that 2 children will not have a seat and 5 children will have a seat when the music stops.

24*21= 504
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 24 Mar 2017, 20:56
first pick 5 children out of 7,, that is 7C5 = 21

then ciricular arragement for 5 chairs = 4!

therefore 21 * 4! = 504

ans D
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 27 Mar 2017, 17:29
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Bunuel wrote:
Seven children are playing musical chairs. When the music stops, in how many ways can five of the seven children be arranged in a circle of five chairs?

A. 21
B. 42
C. 360
D. 504
E. 2520


Since there are 7 children and 5 chairs, we first need to select 5 children from 7.

7C5 = 7!/(5! X 2!) = (7 x 6)/2 = 21 ways

Next we need to arrange those 5 children in a circle, which can be done in (5 - 1)! = 4! = 24 ways.

Thus, 5 of the 7 children can be arranged in a circle of 5 chairs in 21 x 24 = 504 ways.

Answer: D
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 22 Apr 2017, 20:01
Bunuel wrote:
Seven children are playing musical chairs. When the music stops, in how many ways can five of the seven children be arranged in a circle of five chairs?

A. 21
B. 42
C. 360
D. 504
E. 2520


Considering the selection of 5 children from 7 alone

7c5= 21 arrangements

Considering the possible arrangments from 5 children

(n-1)!= (5-1)!

Multiply the two

7c5 x (n-1)!
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 26 Apr 2018, 10:50
Bunuel wrote:
Seven children are playing musical chairs. When the music stops, in how many ways can five of the seven children be arranged in a circle of five chairs?

A. 21
B. 42
C. 360
D. 504
E. 2520



7C5 * (5-1)! = 21 * 24 = 504, hence D is the answer
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Re: Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 24 Sep 2018, 23:36
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Quote:
This combinatorics problem can be approached in multiple ways depending on the story that one tells to describe it. One approach is to treat it as a pure linear permutation and then divide out duplicate orders to account for the circular nature of the arrangement. Another approach is to separate the problem into a combination to choose the people and then a circular arrangement to order them.

Taking the former approach, the problem is "yes order" and "no replacement," so it's a permutation. There are 7 choices for the first chair, 6 for the second, 5 for the third, 4 for the fourth, and 3 for the fifth. As a linear permutation, the answer would be 7∗6∗5∗4∗3

However, this permutation is circular, which means that we could have started placing children from any of the five seats. In other words, from a circular perspective, there is no real difference between the arrangement ABCDE and the arrangements BCDEA, CDEAB, DEABC, and EABCD. In fact, we have overcounted by this same factor of five in each of our orderings. So the real answer must be (7∗6∗5∗4∗3)/5 =7∗6∗4∗3=504
(Note that we could estimate, examine the units digits, and look at the answer choices to avoid having to put too much effort into this calculation; it's 42∗12, so it's greater than 400 and it ends in a 4) This is D.


Quote:
Another option is to treat the problem first as a combination. We have seven children, but only five will get seats. Determining which children get to sit is a "no order," "no replacement" proposition, so it's a combination, and we can use the formula

nCk=n!/(n−k)!∗k!
In this formula, n represents the number of items from which we will choose and k represents the number of picks that will be made. With n=7 and k=5, the formula gives

7C5=7!/2!∗5!=7∗6/2=21
Having chosen the five children who will sit, we turn to arranging them in a circle. We can use the rule that the first item is "free" in a circular arrangement. (It doesn't really matter where the first child sits, in terms of the relative positions of the children.) The arrangement can therefore be calculated as 1∗4∗3∗2∗1=24
Finally, since we must choose the children and arrange them, we want 21∗24=504
(again, estimate and/or look at units digits to save effort). The answer is D.

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Seven children are playing musical chairs. When the music stops, in ho  [#permalink]

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New post 25 Sep 2018, 08:11
Bunuel wrote:
Seven children are playing musical chairs. When the music stops, in how many ways can five of the seven children be arranged in a circle of five chairs?

A. 21
B. 42
C. 360
D. 504
E. 2520


The number of ways we can select 5 children of 7 children are in:
7C5 = 21 ways

Now we have to arrange these 5 children in a circle. [ NOTE: We can arrange 'n' number of people in a circle in (n-1)! ways ]
(5 - 1)! = 4! = 24 ways.

Therefore, the number of ways in which five of the seven children be arranged in a circle of five chairs is 21 x 24 = 504 ways

Answer is D
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Seven children are playing musical chairs. When the music stops, in ho &nbs [#permalink] 25 Sep 2018, 08:11
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