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Jill bought 6 glasses for her kitchen - white, red, black, grey, yello

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Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 29 Jan 2018, 00:16
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Jill bought 6 glasses for her kitchen - white, red, black, grey, yellow, and blue - and would like to display 3 of them on the shelf next to each other. If she decides that a red and a blue glass cannot be displayed together at the same time, in how many different ways can Jill arrange the glasses?

A. 24
B. 48
C. 96
D. 120
E. 720

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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 29 Jan 2018, 00:39
Answer is B

First let us take Red in the first position. So we will be left with two places. Now we can’t use red so the remaining number of possibilities will be 4*3=12 ways

In the same way replace red with blue and we get 12 more ways

Now let us eliminate both red and blue and we will be left with 4*3*2 = 24 ways

So total will be 12+12+24 = 48 ways


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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 29 Jan 2018, 05:08
imo answer would be: 96

No of ways Jill can arrange the glasses = total no of ways for selecting and arranging them - total no of ways when red and blue are together

N.o.w= 6P3 - 4*3*2 = 120-24= 96
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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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Total no of ways of arranging the glasses =6x5x4= 120ways
Total no of ways in which red and blue glasses can be displayed together =3!x4=24
So total no of ways in which red and blue are not together =120-24=96
So answer should be C

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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 29 Jan 2018, 10:49
The total ways of arranging the glasses are
1. Arranging 4 from white, black, grey, and yellow = 4P3
2. Selecting any 2 from white, black, grey, and yellow and 1 from red or blue = 2* 4C2*3!
= 4P3 + 2*4C2*3!= 96
Answer = C
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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 30 Jan 2018, 00:28
why is it that "Total no of ways of arranging the glasses =6x5x4= 120ways
Total no of ways in which red and blue glasses can be displayed together =3!x4=24
So total no of ways in which red and blue are not together =120-24=96
So answer should be C "

i totally understand the way of the calculation except for the part where the number of red and blue glasses are subtracted

how do you build the Formular " 3!x4=24"
how do you come up with 3 and 4?
thank you
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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 30 Jan 2018, 05:28
There are 3! Ways in which 3 glasses can be arranged and there are 4 ways in which the 3rd glass can be selected. So 3!x4

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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello [#permalink]

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New post 31 Jan 2018, 16:57
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Bunuel wrote:
Jill bought 6 glasses for her kitchen - white, red, black, grey, yellow, and blue - and would like to display 3 of them on the shelf next to each other. If she decides that a red and a blue glass cannot be displayed together at the same time, in how many different ways can Jill arrange the glasses?

A. 24
B. 48
C. 96
D. 120
E. 720



We can use the equation:

Number of ways with red and blue glasses not together = total number of arrangements - red and blue glasses together.

Since the order of the glasses is important, we use permutations. Thus, the total number of arrangements is:

6P3 = 6!/(6-3)! = 6!/3! = 6 x 5 x 4 = 120

Since there 4 ways to choose a glass (other than red and blue) along with the red and blue glasses, and once three glasses are picked, there are 3! ways to arrange them, the number of arrangements with red and blue glasses together in a display is:

4 x 3! = 4 x 6 = 24

Thus, the number of ways with red and blue glass not together in a display is:

120 - 24 = 96

Answer: C
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Re: Jill bought 6 glasses for her kitchen - white, red, black, grey, yello   [#permalink] 31 Jan 2018, 16:57
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