On Tuesday, Kramer purchases exactly 3 new shirts, 2 new swe

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Mar 13, 2019 8:38 pm

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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

Source: Manhattan Prep

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Source: Beat The GMAT — Problem Solving | Wed Mar 13, 2019 8:38 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Wed Mar 13, 2019 8:58 pm

BTGmoderatorDC wrote: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

Source: Manhattan Prep

So, Kramer will start wearing outfits, starting Wednesday and this will go on. We have to find out the day when it is not possible for him to wear a new combination of shirt, sweater, and hat.

He has exactly 3 new shirts, 2 new sweaters, and 4 new hats; thus, the total number of combinations possible = 3*2*4 = 24.

So, 25th day from Wednesday, inclusive, he will have to repeat one of the combinations. There are 3 full weeks in 24 days, so we in 3*7 = 21 days, he will come to Tuesday; thus, the 4th day (= 25 - 21) starting Wednesday is Saturday.

On Saturday, he will have to repeat one of the combinations.

The correct answer: E

Hope this helps!

-Jay
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swerve: Legendary Member; Posts: 2548; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Thu Mar 14, 2019 9:10 am

Shirt combinations $= \frac{3!}{1*(3-1)!} = \frac{3*2*1}{1*2*1} = 3$

Sweater combinations $= \frac{2!}{1*(2-1)!} = \frac{2*1}{1*1} = 2$

Hat combinations $= \frac{4!}{1!*(4-1)!} = \frac{4*3*2*1}{1*3*2*1} = 4$

Total Combinations = Shirt Combos * Sweater Combos * Hat Combos
$3*2*4=24$ combinations

$24 / 7$ (days of the week) $=$ remainder of $3$

Purchased on Tuesday, 3rd day is Friday. Next day, which he won't have a new combination for, is Saturday.

Regards!

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8156; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon Mar 18, 2019 5:33 pm

BTGmoderatorDC wrote: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

Source: Manhattan Prep

There are 3 x 2 x 4 = 24 different combinations of outfits. Since he starts to wear the first combination on Wednesday, he will wear the 8th, 15th and 22nd combinations on subsequent Wednesdays. So he will wear the 23rd on Thursday and the 24th on Friday. Therefore, we see that he can no longer wear a different outfit on Saturday since he has exhausted all 24 combinations of outfits on Friday.

Answer: E

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