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The number of ways in which 3 paintings can be arranged in

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The number of ways in which 3 paintings can be arranged in [#permalink] New post 16 Dec 2007, 11:04
The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.

Here, the answer is 36, but I don't have the explanation to this one. can someone show me? thanks
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Re: combinations [#permalink] New post 16 Dec 2007, 11:55
tarek99 wrote:
The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.

Here, the answer is 36, but I don't have the explanation to this one. can someone show me? thanks


= 1 x 4 x 3 x (3)
= 4x3x3
= 36
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 [#permalink] New post 16 Dec 2007, 11:58
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N=(5-1)C(3-1)*3P3=(4*3/2)*(3*2)=36
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Re: combinations [#permalink] New post 16 Dec 2007, 14:37
GMAT TIGER wrote:
tarek99 wrote:
The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.

Here, the answer is 36, but I don't have the explanation to this one. can someone show me? thanks


= 1 x 4 x 3 x (3)
= 4x3x3
= 36



Can you elaborate on this please?
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Re: combinations [#permalink] New post 16 Dec 2007, 21:34
GMATBLACKBELT wrote:
GMAT TIGER wrote:
tarek99 wrote:
The number of ways in which 3 paintings can be arranged in an exhibition from a set of five, such that one is always included.

Here, the answer is 36, but I don't have the explanation to this one. can someone show me? thanks


= 1 x 4 x 3 x (3)
= 4x3x3
= 36



Can you elaborate on this please?


5 paintings: 1, 2, 3, 4, and 5.
three places: A B C

lets chose 1 that always included.

suppose 1 at place A = 1x4x3
suppose 1 at place B = 4x1x3
suppose 1 at place C = 4x3x1

so total = (1x4x3) + (4x1x3) + (4x3x1) = 1x4x3 (3) = 36
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 [#permalink] New post 16 Dec 2007, 21:39
walker wrote:
N=(5-1)C(3-1)*3P3=(4*3/2)*(3*2)=36


if one is always included, 5C3 becomes 4C2.

4C2 = 4!/2!2! = 4*3/2 = 6

arrangements of 3 items = 3!

3! * 6 = 36
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 [#permalink] New post 17 Dec 2007, 00:22
A B C

1 is always included:

4C2 * 3! (3! 'cause order does not matter)
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 [#permalink] New post 17 Dec 2007, 00:58
thanks guys. but i have a question. what are you logically doing when you multiply 6 by 3!? i don't understand the exact function of the 3! sometimes i see people divide the final number by a factorial and then sometimes i see people multiply. what are we logically doing when we do that? can someone explain?
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 [#permalink] New post 17 Dec 2007, 02:09
tarek99 wrote:
thanks guys. but i have a question. what are you logically doing when you multiply 6 by 3!? i don't understand the exact function of the 3! sometimes i see people divide the final number by a factorial and then sometimes i see people multiply. what are we logically doing when we do that? can someone explain?


The logic here is that, basically, we have 3 chairs where we should put 3 apples (A, B or C). What is the number of such combinations?
1st place: we could put any of 3 apples (A or B or C). We put A (but we could place B or C <-- order does not matter)
2nd place: we left with 2 apples (B or C) - so we put B (but we could put C)
3rd place - we left with only 1 apple

ABC
ACB
BAC
BCB
CAB
CBA

Total number of combinations is 3 * 2 * 1 = 3! = 6

But factorial is easy, try to place 4 apples on for chairs. Hope this helps.

P.S> If order matters, i.e. we need apple A on the first chair, the number of combinations would be: 1(A) * 2 (B or C) * 1 (the one that left after B or C)
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 [#permalink] New post 18 Dec 2007, 12:37
tarek99 wrote:
thanks guys. but i have a question. what are you logically doing when you multiply 6 by 3!? i don't understand the exact function of the 3! sometimes i see people divide the final number by a factorial and then sometimes i see people multiply. what are we logically doing when we do that? can someone explain?


3! implies 3*2*1. This means that we are arranging items (there is a sharp diference between arranging and selecting). we can have 3 items in the first slot, 2 items in the second and only 1 item in the last slot.

you need to look at the formula for combinations and permutations. look at the combination formula. then look at the permuation formula.

the perm formula does not have r!. dividing by r! removes the items that are arranged.
  [#permalink] 18 Dec 2007, 12:37
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