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Father John forms a choir from the church attendants. 30 [#permalink]

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12 Nov 2012, 17:27

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Father John forms a choir from the church attendants. 30 people attend John's church, and the choir has 28 spots available, with one person as the lead singer. How many different combinations does John have?

Now, there are 30 choices for the lead vocal - multiply the number of choices for the choir by 30, because you need a lead AND 27 other members: 30×29×14 choices

Hmmm. This is painful to calculate, but the solution is a number that ends with 0 (why? look at the unit's digit for each of the numbers that are multiplied: 0×9×4 = something that ends with 0).

12180 is the only answer that ends with 0. Hallelujah John!

Not understanding how they came up with 30x29x14 Can anyone explain?

Father John forms a choir from the church attendants. 30 people attend John's church, and the choir has 28 spots available, with one person as the lead singer. How many different combinations does John have? (A) 3005 (B) 4412 (C) 6544 (D) 12180 (E) 24366

First of all, one thing that will make a huge difference is who the lead singer is. We have 30 choices for the lead singer. So count that first. That's where the 30 comes from.

Now, for each choice of lead singer, there are 29 folks remaining. This means, for a choir of 28 counting the lead singer, we need to select 27 of these remaining 27 people. Mathematically, that's

29C27

That would be ugly to calculate --- instead, we'll use the symmetry of Pascal's Triangle to reduce that to 29C27 = 29C2 --- another way to say that is --- choosing the 27 of the 29 are included is entirely equivalent to choosing the two of 29 who will not be excluded and told to go home. That's also why 29C27 = 29C2.

Here, it's useful to know the handy shortcut:

nC2 = n(n-1)/2 (which, incidentally, is also the sum of the first n integers, but that's another story.)

So,

29C2 = (29*28)/2 = 29*14

For each one of the 30 choices for lead singer, we could form 29*14 different choirs to back up that lead singer, for a grand total of

30*29*14

Does all this make sense? That was a lot. Please let me know if you have further questions on any of this.

Re: Father John forms a choir from the church attendants. 30 peo [#permalink]

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12 Nov 2012, 20:53

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you can think about it this way.... you need to form a team of 28 players out of 30... so team combinations will be 30C28.... for any team you selected you need to select one player as a captain..... for a given team any player of the team can be a captain... so 28 teams with different captains (even though team is the same).... so total possible combinations will be 30C28*28 which is 30*29*14 (expand 30C28) = 12180... you don't even need to do any multiplication... 30 has a 0 in units place.... so answer should be having a 0 in units place.... so the only answer is 12180

The basic idea is that "30C28" (read as "thirty choose twenty-eight") is the number of combination of 28 items we could select from a pool of 30 unique items. For simplicity, let's pick a simpler number.

5C3 would be the number of combinations of three I could pick from a pool of five unique items. As it happens, 5C3 = 10. Think about the set {A, B, C, D, E}, which has five unique items. Here is the set of 10 possible combinations of three

A, B, C A, B, D A, B, E A, C, D A, C, E A, D, E B, C, D B, C, E B, D, E C, D, E

Notice, we are counting "combinations" in which order doesn't matter, so ABC and CAB would count as the same combination. When order does matter, that's something called "permutation", which is also discussed in that blog article.

To evaluate a number like 6C2 ----- a) some calculators have nCr as a function, but of course, you won't have that available on the GMAT b) you can use the formula nCr = (n!)/[(r!)((n-r)!)] c) You can use Pascal's Triangle http://en.wikipedia.org/wiki/Pascal%27s_triangle

Writing out to the six row of Pascal's triangle (the top 1 counts as the "zeroth" row), we get

Attachment:

Pascal's triangle, up to n = 6.JPG [ 12.63 KiB | Viewed 1665 times ]

You see, nCr is the rth entry on the nth row of Pascal's triangle. (We have to remember to start counting at zero for both the rows and the entries.)

In the six row, the zeroth entry is 1, the first entry is 6, and the second entry is 15, so 6C2 = 15.

Notice the symmetry ----- 6C4 = 6C2, and more generally, nCr = nC(n-r)

When r = 2, a great shortcut you can use is nC2 = n(n-1)/2 That's very useful for things like 30C2, in which case it would be prohibitive to write out Pascal's triangle all the way.

Re: Father John forms a choir from the church attendants. 30 peo [#permalink]

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14 Nov 2012, 22:49

Amateur wrote:

you can think about it this way.... you need to form a team of 28 players out of 30... so team combinations will be 30C28.... for any team you selected you need to select one player as a captain..... for a given team any player of the team can be a captain... so 28 teams with different captains (even though team is the same).... so total possible combinations will be 30C28*28 which is 30*29*14 (expand 30C28) = 12180... you don't even need to do any multiplication... 30 has a 0 in units place.... so answer should be having a 0 in units place.... so the only answer is 12180

Great! I did the same 28C30* 28, but I wasted time on multiplying 15*29*28 which is ridiculous. _________________

Re: Father John forms a choir from the church attendants. 30 [#permalink]

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14 Jul 2014, 19:26

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Re: Father John forms a choir from the church attendants. 30 [#permalink]

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05 Aug 2015, 13:01

OE:

Pick the lead first - there are 30 choices. Now John has to pick 27 more choir members out of the 29 still available. The order of picking the other members does not matter, since they are not assigned different roles in the choir. Since John can't pick the same person twice, there is no repetition. Use the combinations formula for selecting 27 out of 29, no repetition, not ordered.

Now, there are 30 choices for the lead vocal - multiply the number of choices for the choir by 30, because you need a lead AND 27 other members: 30×29×14 choices

Hmmm. This is painful to calculate, but the solution is a number that ends with 0 (why? look at the unit's digit for each of the numbers that are multiplied: 0×9×4 = something that ends with 0).

12180 is the only answer that ends with 0. Hallelujah John! _________________

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