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A certain car dealership offers its newest model in ten [#permalink]
11 Mar 2013, 22:41

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Difficulty:

5% (low)

Question Stats:

80% (01:48) correct
20% (00:47) wrong based on 106 sessions

A certain car dealership offers its newest model in ten exterior and ten interior colors. If the ten interior colors are identical to the ten exterior colors, and the dealership sells every pair of colors except those that would result in a car with an identically colored interior and exterior, how many different color combinations are possible?

Re: A certain car dealership offers its newest model in ten [#permalink]
12 Mar 2013, 01:07

Expert's post

emmak wrote:

A certain car dealership offers its newest model in ten exterior and ten interior colors. If the ten interior colors are identical to the ten exterior colors, and the dealership sells every pair of colors except those that would result in a car with an identically colored interior and exterior, how many different color combinations are possible?

A. 45 B. 81 C. 90 D. 10!/2! E. 10!

There are 10*10=100 different combinations of interior and exterior, out of which 10 will be with an identically colored interior and exterior (1-1, 2-2, 3-3, ..., 10-10), thus the answer is 100-10=90.

Re: A certain car dealership offers its newest model in ten [#permalink]
08 Nov 2013, 00:23

Bunuel wrote:

emmak wrote:

There are 10*10=100 different combinations of interior and exterior, out of which 10 will be with an identically colored interior and exterior (1-1, 2-2, 3-3, ..., 10-10), thus the answer is 100-10=90.

Could you also solve it by 10!/8! --> 10x9 --> 90 ?

Re: A certain car dealership offers its newest model in ten [#permalink]
08 Nov 2013, 00:27

Expert's post

vince44 wrote:

Bunuel wrote:

emmak wrote:

There are 10*10=100 different combinations of interior and exterior, out of which 10 will be with an identically colored interior and exterior (1-1, 2-2, 3-3, ..., 10-10), thus the answer is 100-10=90.

Could you also solve it by 10!/8! --> 10x9 --> 90 ?

I don't think so. What's your logic behind 10! and 8!? _________________

Re: A certain car dealership offers its newest model in ten [#permalink]
28 Nov 2014, 04:14

Expert's post

Ralphcuisak wrote:

Bunuel, I solved it by 10C2 * 2!(as order is not required)..is that correct?

Yes. 10C2 gives the number of different 2-color combinations possible out of 10 and multiplying by 2 takes into the account that interior can be of one color and the exterior of another or vise-versa. _________________

Re: A certain car dealership offers its newest model in ten [#permalink]
29 Nov 2014, 04:18

The question says there are 10 different colors. Since interior and exterior are to be painted with different colors, the order is important while counting the color combination. Therefore, the number of possible ways can be found out by permutation of 2 colors out of 10 colors. 10P2 = 10!/(10-2)! = 10!/8! = 90 [10P2 = 10C2*2!]

Re: A certain car dealership offers its newest model in ten [#permalink]
10 Jan 2015, 04:16

I lined up the 10 different colours for Exterion and Interior like this:

E: A B C D E F G H I J I: A B C D E F G H I J

I calculated the number of combinations: 10*10= 100 combinations. But, from these 100 we need to delete sth: the same colours.

There are 10 possible pairs of coulours that will be the same. So, I deleted these 10 pairs from 100, resulting in 90. Mind here, that the 10 do not result in 20 colours, because 1 pair is one unit that is not allowed. So, 10 units are not allowed (surprised that 80 wasn't among the answer options to confuse test takers...).

One question: I was thinking of the answer given by vince44. Why is it not possible to solve it by 10!/8!. Could this mean: out of the 10 you deduct the same pairs? 8 will be different, but 2 (one pair) will be the same. So, we deduct this one. Because the 10 are pairs of different colours and 2 are one pair of same colours. So, by the method of anagramming you would have ABCDEFGHSS, where S is the Same colours; which would be 10!/8!...?

Really not sure though. But, reading this solution it had me thinking...

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