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old_dream_1976
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7 Carpenters 9 masons Form a team of 3 with atleast one 02 Jan 2006, 13:19
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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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ywilfred
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7 Carpenters 9 masons Form a team of 3 with atleast one 02 Jan 2006, 18:10
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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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hkm_gmat
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7 Carpenters 9 masons Form a team of 3 with atleast one 02 Jan 2006, 18:36
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wilfred is the pro, but i am getting 441

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

Where am i missing the x2 ??
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7 Carpenters 9 masons Form a team of 3 with atleast one 02 Jan 2006, 20:16
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[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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7 Carpenters 9 masons Form a team of 3 with atleast one 02 Jan 2006, 20:34
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hkm_gmat
wilfred is the pro, but i am getting 441

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

Where am i missing the x2 ??
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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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ywilfred
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7 Carpenters 9 masons Form a team of 3 with atleast one 03 Jan 2006, 00:24
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duttsit
hkm_gmat
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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my bad... :oops: it's a combination question and not a permutation, so the answer should be 441.
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7 Carpenters 9 masons Form a team of 3 with atleast one 04 Jan 2006, 09:48
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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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7 Carpenters 9 masons Form a team of 3 with atleast one Updated on: 19 Jan 2006, 10:30
Originally posted by rlevochkin on 19 Jan 2006, 10:22.
Last edited by rlevochkin on 19 Jan 2006, 10:30, edited 1 time in total.
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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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7 Carpenters 9 masons Form a team of 3 with atleast one 19 Jan 2006, 10:28
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First choose one from each group, then choose the third one from the remaining 14 people.

7C1*9C1*14C1 =882
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chiragr
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7 Carpenters 9 masons Form a team of 3 with atleast one 19 Jan 2006, 10:34
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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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chiragr
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7 Carpenters 9 masons Form a team of 3 with atleast one 19 Jan 2006, 10:38
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changing sides... to 441 need to divide 882/2! for combination!



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