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stupid pants

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by diebeatsthegmat » Sun Sep 19, 2010 1:30 pm
Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn't wear 2 specific shirts with 3 specific pants?

a) 41.
b) 66.
c) 36.
d) 70.
e) 56.

[spoiler]how to solve this problem? i dont understand its OE
There are (8 x 9) 72 possibilities of shirts + pants. (2 x 3) 6 Of the combinations are not allowed. Therefore, only (72 - 6) 66 combinations are possible.
is it not right that there will be 6 shirts and 6 pants left because barbara doesnt wear 2 special shrts and 3 special pants?
and so 6*6=36?[/spoiler]
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by shaw3257 » Sun Sep 19, 2010 3:26 pm
diebeatsthegmat wrote:Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn't wear 2 specific shirts with 3 specific pants?

a) 41.
b) 66.
c) 36.
d) 70.
e) 56.
So we have 8 shirts that go with 9 pants. Therefore the maximum number of combinations we can have is:

8 * 9 = 72

Now it is told to us that two of the specific shirts can't go with 3 specific pants. Lets call these shirts S1 and S2, and lets call these pants P1, P2, and P3.

Listing out all the possible ways the shirts can combine with the pants we get:

S1 P1
S1 P2
S1 P3
S2 P1
S2 P2
S3 P3

6 ways

Now subtract the this number from the total possible ways can combine 8 shirts with 9 pants and we get:

72-6=66

pick B
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by goyalsau » Mon Sep 20, 2010 5:18 am
shaw3257 wrote:
diebeatsthegmat wrote:Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn't wear 2 specific shirts with 3 specific pants?

a) 41.
b) 66.
c) 36.
d) 70.
e) 56.
So we have 8 shirts that go with 9 pants. Therefore the maximum number of combinations we can have is:

8 * 9 = 72

Now it is told to us that two of the specific shirts can't go with 3 specific pants. Lets call these shirts S1 and S2, and lets call these pants P1, P2, and P3.

Listing out all the possible ways the shirts can combine with the pants we get:

S1 P1
S1 P2
S1 P3
S2 P1
S2 P2
S3 P3

6 ways

Now subtract the this number from the total possible ways can combine 8 shirts with 9 pants and we get:

72-6=66

pick B
Good work Man,
can we solve it with combination formula...
I am not sure i tried not able to find the answer ,
LOOKED at your explanation and you put it in a very simple and easy way,
But can it be done with combination.
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by shaw3257 » Mon Sep 20, 2010 1:37 pm
goyalsau wrote:
Good work Man,
can we solve it with combination formula...
I am not sure i tried not able to find the answer ,
LOOKED at your explanation and you put it in a very simple and easy way,
But can it be done with combination.
We were secretly using the combination formula for above, but I'll show it differently:

Number of ways 8 shirts can go with 9 pants:

8C1 * 9C1 = 72

Number of ways 2 shirts can go with 3 pants:

2C1 * 3C1 = 6

Number of ways to select a pair:

72-6 = 66
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by kaushiksin » Mon Sep 20, 2010 5:52 pm
Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn't wear 2 specific shirts with 3 specific pants?

Answer


Possible clothing combinations = Total combinations - Combinations in which specific shirts & pants can't be used

Total combinations = 8*9 = 72

Combinations, where specific shirts & pants can't be used = {(P1,S1), (P1,S2), P1,S3), (P2,S1), (P2,S2), (P2,S3)} = 6

Hence, Possible clothing combinations = 72-6 = 66

The answer will be 66.
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