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# John has four ties, twelve shirts, and three belts. If each day he we

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Math Expert
Joined: 02 Sep 2009
Posts: 47879
John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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19 Jul 2018, 21:42
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Difficulty:

5% (low)

Question Stats:

98% (01:02) correct 2% (02:38) wrong based on 52 sessions

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John has four ties, twelve shirts, and three belts. If each day he wears exactly one tie, one shirt, and one belt, what is the maximum number of days he can go without repeating a particular combination?

A. 12
B. 21
C. 84
D. 144
E. 108

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John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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19 Jul 2018, 21:56
John has four ties, twelve shirts, and three belts.

Everyday he has to make 3 choices

Selecting a tie AND selecting a shirt AND selecting a belt

No of ways to select a tie among 4 ties = $$4c_ 1$$ = 4 ways

No of ways to select a shirt among 12 shirts = $$12c_ 1$$ = 12 ways

No of ways to select a belt among 3 belts = $$3c_ 1$$ = 3 ways

=> Total number of ways

=> 4 ways to select a tie AND 12 ways to select a shirt AND 3 ways to select a belt

=> 4 * 12 * 3

= 144

Hence option D
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Director
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Re: John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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20 Jul 2018, 03:14
Bunuel wrote:
John has four ties, twelve shirts, and three belts. If each day he wears exactly one tie, one shirt, and one belt, what is the maximum number of days he can go without repeating a particular combination?

Number of ways to select a Tie = 4C1 = 4
Number of ways to select a Shirt = 12C1 = 12
Number of ways to select a Belt = 3C1 = 3

Therefore,
Possible Number of ways combined = 4*12*3 = 12*12 = 144

Hence, D.
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Re: John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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20 Jul 2018, 03:30
Why use combinations? Simply multiply the number of items.

For each of the four ties, you can choose among 12 shirts = 4 x 12 possible combinations.

For each of these combinations, you can choose among three belts so le't multiply the previous number by 3 = 4 x 12 x 3.
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Re: John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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20 Jul 2018, 03:35
yoannnesme wrote:
Why use combinations? Simply multiply the number of items.

For each of the four ties, you can choose among 12 shirts = 4 x 12 possible combinations.

For each of these combinations, you can choose among three belts so le't multiply the previous number by 3 = 4 x 12 x 3.

yoannnesme
Combination is not required as per the stem here, its just that - Old habits die hard.
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Re: John has four ties, twelve shirts, and three belts. If each day he we  [#permalink]

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23 Jul 2018, 18:25
Bunuel wrote:
John has four ties, twelve shirts, and three belts. If each day he wears exactly one tie, one shirt, and one belt, what is the maximum number of days he can go without repeating a particular combination?

A. 12
B. 21
C. 84
D. 144
E. 108

The number of tie-shirt-belt combinations is 4 x 12 x 3 = 144.

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Re: John has four ties, twelve shirts, and three belts. If each day he we &nbs [#permalink] 23 Jul 2018, 18:25
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