Fernando purchased a university meal plan that allows him to

Question

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?

(A) 20

(B) 24

(C) 40

(D) 100

(E) 120

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5C3 × 4C3

(A) 20

(B) 24

(C) 40

(D) 100

(E) 120

5C3 × 4C3

argha · Answer

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

Regards

Argha

Bunuel · Answer

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.

Asifpirlo · Answer

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

GMAT40 · Answer

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

Archit3110 · Answer

total possible ways 
5c3 for lunches and 4c3 for dinners 
5c3*4c3 = 10*4 ; 40
IMO C

Rebaz · Answer

Bunuel

Can you please explain how 5C3*4C3= 40?

My solution:

5C3= 5*4*3*2*1 / 3*2*1 = 20 and(Mulitply) by 4C3= 4*3*2*1 / 3*2*1 = 4

Thus: 20 * 4 = 80.

Thanks in advance!

Bunuel · Answer

5C3 = 5!/(3!2!) = 10, not 20.

Rebaz · Answer

Of course it is 10.  Silly me again!

I was so frustrated why it was 10 and not 20. I think a need a cup of coffee!

Thank you very much for your quick response, and you are a HERO!

architkap · Answer

can someone explain why 9c6 doesnt work here? 9 total meals (5 Lunches + 4 dinners) and we have to choose 6.

Bunuel · Answer

9C6 counts every way to choose 6 meal slots from the 9 available slots. However, it also includes invalid selections, such as 4 lunches and 2 dinners or 5 lunches and 1 dinner.

Fernando must choose  exactly 3 of the 5 lunch slots and exactly 3 of the 4 dinner slots , so the choices must be counted separately:

5C3 * 4C3.

egmat · Answer

Hi architkap,

Good instinct to look for a one-shot count, but here's the snag:  lunches and dinners aren't interchangeable slots , so they can't go into one pool.

Look back at Argha's and Bunuel's setup: the plan forces  exactly 3 lunches  and  exactly 3 dinners . When you do  9C6 , you're just grabbing any  6  of the  9  slots - with no rule that  3  of them must be lunches and  3  must be dinners.

That's the constraint your method quietly drops. For example,  9C6  happily counts a pick like  5 lunches + 1 dinner , or  2 lunches + 4 dinners . Neither is a legal meal plan, but they're all inside your  9C6  count. So you'd be over-counting with a bunch of invalid combos.

The fix is to keep the two choices  separate , because they're independent:

- Lunches: choose  3  of the  5  weekdays -  5C3 = 10 
- Dinners: choose  3  of the  4  allowed days (no Friday) -  4C3 = 4 
- Multiply, since the two are independent -  10 × 4 = 40

Quick way to feel the difference:  shrink it down. Say you must pick  1  fruit and  1  drink from {apple, banana} and {tea, coffee}. The right count is  2 × 2 = 4  valid meals. But if you dumped all  4  items in one bag and did  4C2 = 6 , you'd also count "apple + banana" and "tea + coffee" - picks that break the 'one of each' rule. That's exactly what  9C6  does to your meal plan.

Whenever a problem fixes  how many from each separate group , count each group on its own and multiply - don't merge them into one pool.

Answer: C

Fernando purchased a university meal plan that allows him to

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