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Father John forms a choir from the church attendants. 30

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

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New post 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?

A. 3005
B. 4412
C. 6544
D. 12180
E. 24366

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

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New post 12 Nov 2012, 20:00
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anon1 wrote:
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

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

Hi, there. I'm happy to help with this. :-)

This problem involves the Fundamental Principle of Counting. You may find these blogs helpful for background.
http://magoosh.com/gmat/2012/gmat-quant-how-to-count/
http://magoosh.com/gmat/2012/gmat-permu ... binations/

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.

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

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

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New post 12 Nov 2012, 21:17
I'm sorry what is the C standing for when you guys write something like 30C28 or 29C27 ?
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Re: Father John forms a choir from the church attendants. 30 peo  [#permalink]

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New post 13 Nov 2012, 03:57
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Re: Father John forms a choir from the church attendants. 30 peo  [#permalink]

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New post 13 Nov 2012, 12:21
anon1 wrote:
I'm sorry what is the C standing for when you guys write something like 30C28 or 29C27 ?

I'll also suggest this blog which discusses the idea.
http://magoosh.com/gmat/2012/gmat-permu ... binations/

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
Pascal's triangle, up to n = 6.JPG [ 12.63 KiB | Viewed 3896 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.

Does all this make sense?

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

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

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New post 05 Aug 2015, 13:01
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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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Re: Father John forms a choir from the church attendants. 30  [#permalink]

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New post 06 Aug 2015, 07:09
Only D is a multiple of 3 and 10.

You don't have to make any multiplication, the "lead singer" kills the question...
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Re: Father John forms a choir from the church attendants. 30  [#permalink]

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New post 10 Jul 2016, 08:17
my thought process: 30(29*28...*3*2)/2 = 15 times even. The last digit has to be 0. So, picked D.
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Re: Father John forms a choir from the church attendants. 30  [#permalink]

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New post 10 Jul 2016, 14:00
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Number of combinations of 28 attendees from 30 (30C28): 30*29/2 = 435

Number of combination of 1 lead singer out of the 28 selected attendees (28): 435*28 = 12,180 (D)
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Father John forms a choir from the church attendants. 30  [#permalink]

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New post 02 Feb 2019, 15:03
Hi All,

We're told that 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. We're asked for the number of different combinations for the choir. This question involves the Combination Formula, but it also includes a Number Property that can help you to avoid almost all of the 'math' involved:

For the lead singer, we have 30 people to choose from --> 30 possibilities.

Once the lead singer is chosen, we have 29 people for the remaining 27 spots....
29c27 = (29!)/(27!)(2!) = (29)(28)/2 = (29)(14)

Thus, the total number of possible choirs is (30)(29)(14). Rather than do all of that math though, notice how we will be multiplying by 30; this means that the product will END in a '0'... and there's only one answer that fits that pattern.

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

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New post 06 Feb 2019, 19:36
anon1 wrote:
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. 24366b


We can have any one of 30 people vie for the lead singer spot, and the remaining 29 people vie for the remaining 27 spots. Therefore, we can have a total of

30C1 x 29C27 = 30 x 29C2 = 30 x (29 x 28)/2 = 12,180 combinations

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

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New post 16 Sep 2019, 04:11
30C1 * 28C1 = 15*29*28=15(30-1)(30-2)=15(900-90+2)=15*812=12,180
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Re: Father John forms a choir from the church attendants. 30   [#permalink] 16 Sep 2019, 04:11
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