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Re: James is selecting CDs to take with him on his trip. James will selec [#permalink]
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chetan2u wrote:
Hi ,

The equation is something which we all will be able to get but effort would be in solving it at the earliest..
So 10 of one kind, 12 of another and 15 of two other kinds will give us 10*12*15*15 combinations..
What will be 10*12*15*15..
Should we get into multiplication or there is some way out..
Look at 0s you will have..

10 will give you one..
Each 15 will give you another 0, when multiplied by 2.. so another 2..
Total 0s .. 1+2=3..
Only E has 3 0s..
Ans E


Hey Chetan,

I agree with you. Instead of multiplying all the numbers, once should look for a simpler method to get the answer quickly.

Here is what I did -
    • After getting \(10 * 12 * 15 * 15\)
    • I wrote it as \(120 * 225\).
    • Now, if the multiplication would have been \(100 * 225\), we would have got \(22500\).
    • But then we have \(120\) instead of \(100\), therefore the answer has to be more than \(22500\) and there is only one option which is more than \(22500\),
    that is option E, thus I came to the conclusion that would be the answer :)


Thanks,
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Re: James is selecting CDs to take with him on his trip. James will selec [#permalink]
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Bunuel wrote:
James is selecting CDs to take with him on his trip. James will select 1 jazz disc, 1 pop, 1 country and western and 1 blues. If he has 12 jazz, 10 pop and 15 each of country and western and blues, then how many possible different combinations of CDs could James take on his trip?

A. 52
B. 1,800
C. 6,750
D. 22,500
E. 27,000


We need to determine the number of ways to select 1 jazz CD from 12, 1 pop CD from 10, 1 country and western CD from 15, and 1 blues CD from 15.

Since we are selecting 1 of each, there are:

12C1 x 10C1 x 15C1 x 15C1 = 12 x 10 x 15 x 15 = 27,000 ways to select the CDs.

Answer: E
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Re: James is selecting CDs to take with him on his trip. James will selec [#permalink]
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Re: James is selecting CDs to take with him on his trip. James will selec [#permalink]
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