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Combination problem

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Combination problem

by [email protected] » Mon Aug 04, 2008 12:55 pm
Anyone know the - quickest way to solve this?

On Tuesday, Kramer purchases exactly 3 new shirts, 2 new sweaters, and 4 new hats, On the following day and each subsequent day thereafter, Kramer wears one of his new shirts together with one of his new sweaters and one of his new hats. Kramer avoids wearing the exact same combination of shirt, sweater, and hat for as long as possible. On which day is this no longer possible?

A)Tuesday
B)Wednesday
C)Thursday
D)Friday
E)Saturday

OA - E
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Re: Combination problem

by beeparoo » Mon Aug 04, 2008 1:11 pm
[email protected] wrote:Anyone know the - quickest way to solve this?

On Tuesday, Kramer purchases exactly 3 new shirts, 2 new sweaters, and 4 new hats, On the following day and each subsequent day thereafter, Kramer wears one of his new shirts together with one of his new sweaters and one of his new hats. Kramer avoids wearing the exact same combination of shirt, sweater, and hat for as long as possible. On which day is this no longer possible?

A)Tuesday
B)Wednesday
C)Thursday
D)Friday
E)Saturday

OA - E
First, this isn't a combinations problem. This is a counting problem.

The total number of unique outfits is 3 x 2 x 4 = 24.

Since Kramer begins sporting his new wardrobe on Thursday, 3 weeks and 3 days lapse until he runs out of unique outfits. 3 days after Thursday is Saturday.

Finally, please use the spoiler tags to shield the OA.
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by parallel_chase » Mon Aug 04, 2008 1:41 pm
There are two mistakes in "beeparoo's" post. Though she still manages to get the right answer. For which she rightly mentioned about the answer choices to have spolier tags.

The question asks On which day is this no longer possible?

The answer is Saturday indeed. 3*2*4=24 combination.

He will start to wear the first combination on Wednesday.

Every Tuesday he will complete the cycle of 7 days.

Therefore, on Saturday, he would have to repeat his any of the combination.
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by beeparoo » Mon Aug 04, 2008 2:01 pm
parallel_chase wrote:There are two mistakes in "beeparoo's" post. Though she still manages to get the right answer. For which she rightly mentioned about the answer choices to have spolier tags.

The question asks On which day is this no longer possible?

The answer is Saturday indeed. 3*2*4=24 combination.

He will start to wear the first combination on Wednesday. D'oh! That was a dumb error

Every Tuesday he will complete the cycle of 7 days.

Therefore, on Saturday, he would have to repeat his any of the combination. Yes - true again.
Thanks for the corrections, parallel_chase. It's all in the details, details, details!

P.S. I got the right answer, in spite of my explanation, because I just jot down chicken scratch calculations on paper while keeping a slight mental consciousness of other details. I probably make too many careless errors this way... :roll:
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by canuckclint » Fri Oct 31, 2008 3:33 pm
He starts wearing these on wednesday so
tuesday will be his 7th combination or multiple of 7.

Tuesday - 21st combination
Wednesday - 22nd combination
Thursday - 23rd
Friday - 24th
Sat - No more left
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A matter of framing

by simba12123 » Sat Nov 01, 2008 9:15 am
Regardless of calling this question combinations or counting approach, I think the most critical part of this very easy and very difficult question is a matter of framing. To frame the correct approach is the only real hurdle on this question. Applause to all great minds on this question. Any recommendations as to where to get some framing exercises?

p.s. IMO I feel that this question was a counting question.
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by cramya » Sun Nov 02, 2008 12:59 pm
3*2*4 = 24 ways

24/3!(order does not matter since its the same set of 3) = 24/ 3*2 = 4

Tue,Wed,Thu,Fri possible On Sat it wont be a unique pick amongst all three
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by cramya » Sun Nov 02, 2008 1:02 pm
Please ignore my post. I messed it up big time.

Parallel chase explantion and canuckclint got it right in their explanations.
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