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Fernando purchased a university meal plan that allows him to

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Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 13 Aug 2013, 00:05
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Fernando purchased a university meal plan that allows him to have a total of 3 lunches and 3 dinners per week. If the cafeteria is closed on weekends and Fernando always goes home for a dinner on Friday nights, how many options does he have to allocate his meals?

Sorry I dont have the answer choices, but the correct answer is
5C3 × 4C3
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Re: Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 13 Aug 2013, 00:38
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Hi,

Total number of days in a week=5(excluding weekends). Thus he has 5 lunch options and 5 dinner options to allocate his coupons to.

For lunch, he can divide his 3 lunch coupons in 5C3 ways.

Similarly, for 4 dinner options(excluding Friday), he can divide it in 4C3 ways.

Thus total ways=5C3*4C3



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Argha
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Re: Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 13 Aug 2013, 02:08
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aishu4 wrote:
Fernando purchased a university meal plan that allows him to have a total of 3 lunches and 3 dinners per week. If the cafeteria is closed on weekends and Fernando always goes home for a dinner on Friday nights, how many options does he have to allocate his meals?

Sorry I dont have the answer choices, but the correct answer is
5C3 × 4C3


He can allocate his 3 free lunches on any 3 days from 5 (excluding weekends), so in 5C3 ways.
He can allocate his 3 free dinners on any 3 days from 4 (excluding weekends and Friday), so in 4C3 ways.

Total = 5C3*4C3 ways.

Hope it's clear.
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Re: Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 13 Aug 2013, 03:42
aishu4 wrote:
Fernando purchased a university meal plan that allows him to have a total of 3 lunches and 3 dinners per week. If the cafeteria is closed on weekends and Fernando always goes home for a dinner on Friday nights, how many options does he have to allocate his meals?

Sorry I dont have the answer choices, but the correct answer is
5C3 × 4C3

this problem is from veritas probability book.

Sat ... sun .. mon... tue ...wed...thur...fri
off......off.......L+D... L+D ..L+D...L+D... L

3 L and 3 D available.

5 lunches with 3 lunch packs and 4 dinners with 3 dinner packs = 5C3 × 4C3 = 40
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Re: Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 14 Aug 2013, 19:54
Option only for 3 Lunch and 3 Dinner from below

MO TU WE TH FR
L L L L L
D D D D D

5C3 * 4C3
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Re: Fernando purchased a university meal plan that allows him to  [#permalink]

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New post 12 Apr 2019, 06:19
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Re: Fernando purchased a university meal plan that allows him to   [#permalink] 12 Apr 2019, 06:19
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