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7 Carpenters 9 masons Form a team of 3 with atleast one

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Senior Manager
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7 Carpenters 9 masons Form a team of 3 with atleast one [#permalink] New post 02 Jan 2006, 12:19
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7 Carpenters
9 masons

Form a team of 3 with atleast one carpenter and one mason

how many teams are possible?
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 [#permalink] New post 02 Jan 2006, 17:10
We're told there must be one carpenter and one mason. There are 7 carpenters to pick from for the first carpenter and 9 masons to pick from for the last mason.

The last place can be either a carpeneter or a mason.

Number of ways to pick carpenters = 6
Number of ways to pick mason = 8

# of teams = 7*9*6 + 7*9*8 = 882
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 [#permalink] New post 02 Jan 2006, 17:36
wilfred is the pro, but i am getting 441

7C1*9C2 + 7C2*9C1 = 441

Where am i missing the x2 ??
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 [#permalink] New post 02 Jan 2006, 19:16
[quote="ywilfred"]We're told there must be one carpenter and one mason. There are 7 carpenters to pick from for the first carpenter and 9 masons to pick from for the last mason.

The last place can be either a carpeneter or a mason.

Number of ways to pick carpenters = 6
Number of ways to pick mason = 8

# of teams = 7*9*6 + 7*9*8 = 882[/quote]

You mean 7 X 9 X 14

that would be permutations

what about abc, bac, cab..... 6 such permutations

therefore I think it should be 7x9x14/6

But iam not sure. Can someone please confirm?
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 [#permalink] New post 02 Jan 2006, 19:34
hkm_gmat wrote:
wilfred is the pro, but i am getting 441

7C1*9C2 + 7C2*9C1 = 441

Where am i missing the x2 ??


Agree. We are not concerned about order here(its selection problem not arrangement).

This is same if you divide 882(ywilfred ans) by 2! = 2
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 [#permalink] New post 02 Jan 2006, 23:24
duttsit wrote:
hkm_gmat wrote:
wilfred is the pro, but i am getting 441

7C1*9C2 + 7C2*9C1 = 441

Where am i missing the x2 ??


Agree. We are not concerned about order here(its selection problem not arrangement).

This is same if you divide 882(ywilfred ans) by 2! = 2


my bad... :oops: it's a combination question and not a permutation, so the answer should be 441.
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 [#permalink] New post 04 Jan 2006, 08:48
hey guys, I seem to notice I have problems with permutations problems.

Is there a general logic to thinking about these, when coming up with answers ?

And, this may seem dumb, but how do we know its strictly permutations, and not arrangements ?
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 [#permalink] New post 19 Jan 2006, 09:22
1.no. of ways to form=CCM =7*9*C(6;1)=378

2. no. of ways to form=MMC=7*9*C(8;1)=504

Total no. of ways=378+504=882/!2=441
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Last edited by rlevochkin on 19 Jan 2006, 09:30, edited 1 time in total.
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 [#permalink] New post 19 Jan 2006, 09:28
First choose one from each group, then choose the third one from the remaining 14 people.

7C1*9C1*14C1 =882
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 [#permalink] New post 19 Jan 2006, 09:34
Selcet 1 out of 7 = 7
select 1 out of 9 = 9

Now you have 6+8=14 people left so select one of 14 = 14

so total is
7*9*14 = 882

I think we need OE/OA as clear discrepancy between 441 and 882...
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 [#permalink] New post 19 Jan 2006, 09:38
changing sides... to 441 need to divide 882/2! for combination!
  [#permalink] 19 Jan 2006, 09:38
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