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permutation and combinations

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by kvcpk » Sun Aug 08, 2010 10:10 pm
likithae wrote:In how many ways can a pack of 52 cards be divided in 4 sets, three of them having 17 cards each and fourth just 1 card?


how to solve this problem.please help me ...
1 card can be chosen in 52 ways
From 51 cards, we need to pick 17 cards.
Hence 51C17
Now from 34 cards, pick 17
Hence 34C17
We need not pick any more cards. Because remaining 17 cards will be one set.

All these should happen together. hence multiply:
= 52 * 51C17 *34C17
Order does not matter. Hence divide the result by 3!
= 52 * 51!/(34! * 17!) * 34!/(17!*17!) * 1/3!
= 52 * 51! /(17! * 17! * 17! * 3!)
= 52!/(17!)^3 * 3!

Hope this helps!!
Last edited by kvcpk on Tue Aug 10, 2010 2:44 am, edited 1 time in total.
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by likithae » Tue Aug 10, 2010 2:00 am
kvcpk wrote:
likithae wrote:In how many ways can a pack of 52 cards be divided in 4 sets, three of them having 17 cards each and fourth just 1 card?


how to solve this problem.please help me ...
1 card can be chosen in 52 ways
From 51 cards, we need to pick 17 cards.
Hence 51C17
Now from 34 cards, pick 17
Hence 34C17
From 17, pick 17
Hence 17C17

All these should happen together. hence multiply:
= 52 * 51C17 *34C17 * 17C17
Whats OA?
the OA is 52!/(17!)^3 .3! .but i don't know how to solve it........please help me........
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by kvcpk » Tue Aug 10, 2010 2:45 am
likithae wrote: the OA is 52!/(17!)^3 .3! .but i don't know how to solve it........please help me........
Eidted my above post with the solution. Let me know if you have any queries.
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by sdotcruz » Tue Aug 10, 2010 6:01 am
kvcpk:

Can you tell me how you determined that we should divide by 3? In your post you said that order did not matter so you divided by 3. I am confused because the equations that you used to solve the problem took into account that order did not matter. Did you have to divide by 3 because you can pick each set in three different orders and the order that cards are distributed to the different sets doesn't matter as well? For instance you can distribute 17 cards from the 51 cards to the first set or the second or the third?
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by kvcpk » Tue Aug 10, 2010 7:05 am
sdotcruz wrote:kvcpk:

Can you tell me how you determined that we should divide by 3? In your post you said that order did not matter so you divided by 3. I am confused because the equations that you used to solve the problem took into account that order did not matter. Did you have to divide by 3 because you can pick each set in three different orders and the order that cards are distributed to the different sets doesn't matter as well? For instance you can distribute 17 cards from the 51 cards to the first set or the second or the third?
You are right
I divided the result by 3! Because Order of choosing the sets doesnt matter.
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by anirban_lax » Tue Aug 10, 2010 8:27 am
I thought the problem is pretty easy till I tried understanding the division by 3! part. I don't see where I have considered ordering in the equation when I say - 52*51C17*34C17*1.
Will appreciate if you can you please explain in a little more detail?

Also, I notice that I can choose the set of 1 card first or the set of 17 cards first. Is there any specific reason why you decided to select the set consisting of 1 card, first? Ease of simplification I assume.

Thank you!

-Anirban
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by kvcpk » Tue Aug 10, 2010 8:49 am
anirban_lax wrote:I thought the problem is pretty easy till I tried understanding the division by 3! part. I don't see where I have considered ordering in the equation when I say - 52*51C17*34C17*1.
Will appreciate if you can you please explain in a little more detail?

Also, I notice that I can choose the set of 1 card first or the set of 17 cards first. Is there any specific reason why you decided to select the set consisting of 1 card, first? Ease of simplification I assume.

Thank you!

-Anirban
Yes I did it for ease of simplification.
I can choose any set first. Thats the reason We had to divide by 3!

Hope this helps!!
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by likithae » Wed Aug 11, 2010 12:07 am
kvcpk wrote: 1 card can be chosen in 52 ways
From 51 cards, we need to pick 17 cards.
Hence 51C17
Now from 34 cards, pick 17
Hence 34C17
We need not pick any more cards. Because remaining 17 cards will be one set.

All these should happen together. hence multiply:
= 52 * 51C17 *34C17
Order does not matter. Hence divide the result by 3!
= 52 * 51!/(34! * 17!) * 34!/(17!*17!) * 1/3!
= 52 * 51! /(17! * 17! * 17! * 3!)
= 52!/(17!)^3 * 3!

Hope this helps!!
thank you very much..............
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by Stuart@KaplanGMAT » Wed Aug 11, 2010 12:17 am
kvcpk wrote:
anirban_lax wrote:I thought the problem is pretty easy till I tried understanding the division by 3! part. I don't see where I have considered ordering in the equation when I say - 52*51C17*34C17*1.
Will appreciate if you can you please explain in a little more detail?

Also, I notice that I can choose the set of 1 card first or the set of 17 cards first. Is there any specific reason why you decided to select the set consisting of 1 card, first? Ease of simplification I assume.

Thank you!

-Anirban
Yes I did it for ease of simplification.
I can choose any set first. Thats the reason We had to divide by 3!

Hope this helps!!
Hi!

The reason that you have to divide by 3! is because if you don't, you'll end up counting the same distributions multiple times.

Let's look at a simpler question to illustrate the principle:

How many different ways can the letters ABCDEF be distributed to 3 different groups of 2?

Just using combinations, we have:

6C2 * 4C2 * 2C2 = 15 * 6 * 1 = 90

However, since we don't care about which groups the cards go to, we've created some duplicate distributions.

For example, one distribution is:

AB CD EF

Another distribution is:

CD AB EF

The way we've counted the groups, we've counted each of these as a unique distribution. However, they're identical, so we need to eliminate the ones we've counted multiple times.

Since there are 3 items, there are 3! different ways to arrange them. Accordingly, to count the true number of unique distributions, we need to divide our sum by 3!.

So, the correct answer would be 90/3! = 90/6 = 15.

Similarly, in the question in this thread, we have:

(block of 17 "A") (block of 17 "B") (block of 17 "C").

We've counted:

A B C
A C B
B A C
B C A
C A B
C B A

as 6 different distributions of cards; however, all 6 of those are identical distributions. To avoid all those duplicates, we must divide our sum by 3!.
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by lunarpower » Wed Aug 11, 2010 1:14 am
i know that some of you who follow my posts are going to get sick and tired of hearing me say this, but this is another problem that is absolutely not like anything you are going to see on the test.
it's way too technically intensive -- and also not "unusual" enough. the combinatorics problems that tend to show up on the gmat (above the most basic levels) trend in exactly the opposite directions on both of these counts: they still tend to require no more than elementary calculations, but they are usually "unusual" or "different" in some way.

here's an example of what i'm talking about, from GMAT PREP:
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve the committee?
16, 24, 26, 30, 32
notice that what is by far the easiest way to solve this problem has nothing to do with combination formulas -- you just use consecutive multiplication, and then divide by 3! because order doesn't matter among the members of the committee:
(8 x 6 x 4) / 3! = 32
notice that the numbers are 8, 6, 4, rather than 8, 7, 6, because each choice of a committee member rules out both that member and his/her spouse.

you can solve this problem with combinations, but it's more of a hassle -- you have to use combinations to choose the number of couples, after which you have to use three more consecutive multiplications to select from each couple:
(4c3) x 2 x 2 x 2 = 32

--

in any case, in any problem in which you are required to use two or three combinations -- or, in fact, in any problem involving even one combination with an absurdly large number such as 51 to choose from -- you can rest assured that you have stepped well outside the bounds of anything that you are actually going to see on the test.
Ron has been teaching various standardized tests for 20 years.

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Voit esittää kysymyksiä Ron:lle myös suomeksi

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by likithae » Wed Aug 11, 2010 11:26 pm
Stuart Kovinsky wrote:
kvcpk wrote:
anirban_lax wrote:I thought the problem is pretty easy till I tried understanding the division by 3! part. I don't see where I have considered ordering in the equation when I say - 52*51C17*34C17*1.
Will appreciate if you can you please explain in a little more detail?

Also, I notice that I can choose the set of 1 card first or the set of 17 cards first. Is there any specific reason why you decided to select the set consisting of 1 card, first? Ease of simplification I assume.

Thank you!

-Anirban
Yes I did it for ease of simplification.
I can choose any set first. Thats the reason We had to divide by 3!

Hope this helps!!
Hi!

The reason that you have to divide by 3! is because if you don't, you'll end up counting the same distributions multiple times.

Let's look at a simpler question to illustrate the principle:

How many different ways can the letters ABCDEF be distributed to 3 different groups of 2?

Just using combinations, we have:

6C2 * 4C2 * 2C2 = 15 * 6 * 1 = 90

However, since we don't care about which groups the cards go to, we've created some duplicate distributions.

For example, one distribution is:

AB CD EF

Another distribution is:

CD AB EF

The way we've counted the groups, we've counted each of these as a unique distribution. However, they're identical, so we need to eliminate the ones we've counted multiple times.

Since there are 3 items, there are 3! different ways to arrange them. Accordingly, to count the true number of unique distributions, we need to divide our sum by 3!.

So, the correct answer would be 90/3! = 90/6 = 15.

Similarly, in the question in this thread, we have:

(block of 17 "A") (block of 17 "B") (block of 17 "C").

We've counted:

A B C
A C B
B A C
B C A
C A B
C B A

as 6 different distributions of cards; however, all 6 of those are identical distributions. To avoid all those duplicates, we must divide our sum by 3!.

thank you very much..now i understood how to solve it...
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