Combinatronics - MGMAT Exam question

[Rahul]

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

Can we solve this by ANAGRAM Method? If so please explain.

[rfernandez]

I can't see a way to do it with just one anagram. And the only way I can think of using anagrams at all doesn't really help you much. We can construct anagrams for each of the choices: one for the shirt choice, one for the sweater choice, and one for the hats. But again this doesn't result in much.

Here's what I mean:

For the 3 shirts, you could set up the anagram "YNN", yielding 3!/2! = 3. And this makes sense... there are 3 different ways to choose a shirt. You could repeat this for the sweater ("YN") and for the hats ("YNNN"), but again you'd wind up with 2 and 4, respectively.

From that point, you rely on the counting principle and multiply the possible choices together to give you 3 * 2 * 4 = 24.

Rey

[Guest]

Thanks Rey .. that was helpful!

[rfernandez]

You're welcome!

[themarkac]

Is the answer E

1. Step 1 - Find out how many possible combinations of clothes

3 x 2 x 4 = 24

2. Step 2 - Set up a chart for days of the week

W T F S S M T
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 From saturday on he will have no new combinations

[RonPurewal]

Is the answer E

1. Step 1 - Find out how many possible combinations of clothes

3 x 2 x 4 = 24

2. Step 2 - Set up a chart for days of the week

W T F S S M T
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 From saturday on he will have no new combinations

looks good.

note that the 'list all the numbers' approach works for this problem, because you're only talking about 24 different combinations (24 is a small enough number that it's easy to write out that list in 10 seconds or so). however, if the problem involved hundreds of different combinations, then you'd be in trouble.
in that case, then, here's what you'd do: you'd divide the number by seven, and find the REMAINDER. the remainder would correspond to the #s you wrote in the first row: for example, for day #150, the remainder of 150/7 would be 3. therefore, the 150th combination would fall on a friday, because '3' is friday.

in general, any problem dealing with a periodically repeating sequence can be attacked with remainders in this sort of way. BUT you have the right idea: if the total number of combinations/days/whatever is small enough, then you're probably better off just making a list.