/sci/ - Science & Math

[Anonymous No.8855886]

Everyone that looks at this has a different answer, but I know /sci/ will have the correct answer.

[Anonymous No.8855896]

136 if you cannot combine cupcakes and macarons and 342 if you can.

[Anonymous No.8855900]

would (25 C 18) + (25 C 10) work?

[Anonymous No.8855919]

52 C 27

[Anonymous No.8855946]

>>8855900
>>8855919
No and no. You can only choose 20.

The correct answer is 52 choose 20

The discrepancy comes from the colloquial use of or which excludes the possibility of mixing the two types vs. the logical definition of or where mixing is allowed.

[Anonymous No.8855947]

>>8855946
Meant to say 52 choose 25

[Anonymous No.8856002]

>>8855886
Hello again, discrete math anon. Your hw help baitposting has come so far!

[Anonymous No.8856014]

>>8855947
Or equivalently 52 choose 27 :^)

[Anonymous No.8856043]

>>8856002
actually, first thread I've made here and actually curious of the correct answer since no one seems to know it

[Anonymous No.8856045]

>>8856043

These >>8855947 >>8856014
are both the correct answer.

[Anonymous No.8856051]

>>8856045
>>8856045
that was actually my first thought, but the dumb broad in class said i was wrong. Well anyway, thank you.

[None !niqjediPCA No.8856240]

>>8855886
it's a hell of a lot more than that, guys.

Consider this example for the cupcakes...

25 slots, 18 varieties.

Consider the varieties labeled A through Q.

Example 1: AAAAAAAAAAAAAAAAAAAAAAAAA
Example 2: AAAAAAAAAAAAAAAAAAAAAAAAB
Example 3: AAAAAAAAAAAAAAAAAAAAAAAAC
Example 4: AAAAAAAAAAAAAAAAAAAAAAAAD

etc...

We can quickly see that the possible combination of cupcakes quickly mounts to rediculous numbers.

However, as the problem stats that the ORDER does not matter, the above examples are an over-estimate...

So, a more thourough guage is required, as such.

Example 1
A:25
B:0
C:0
D:0
E:0
F:0
G:0
H:0
I:0
J:0
K:0
L:0
M:0
N:0
O:0
P:0
Q:0

Example 2
A:24
B:1
C:0
D:0
E:0
F:0
G:0
H:0
I:0
J:0
K:0
L:0
M:0
N:0
O:0
P:0
Q:0

Again, the possible combinations are far more numerous than has been previously stated in the thread.

[None !niqjediPCA No.8856246]

>>8856240
cont...

Each variety of cupcakes, as a whole (25 of the same type) gives us 25 possible combinations.

Each variety as a majority (24), with a single other cupcake of all of the other varieties, gives us an additional 24 possibilities.

Each variety as a majority (23), with two cupcakes of a single variety, give us an additional 24 possibilities, etcetera...

Going all the way from A to Q, this gives us 25 +(24*18)=457 possible combinations.

And that is JUST with two different types of cupcakes per pack.

When we consider that one can use all 18 varieties of cupcakes, this multiplies the possibilities far beyond hundreds, into possibly tens of thousands of possible combinations.

[None !niqjediPCA No.8856256]

>>8856246
>Going all the way from A to Q, this gives us 25 +(24*18)=457 possible combinations.

Actually, this should be going from 1 to 12 (as a half a pack of cupcakes) for total possible combinations of one or two flavor of, 18+(12*18)=234 combinations.

Then you get into three combinations, four combinations, five combinations, all the way up to using all 18 possible varieties in a pack, in various combinations.

IT's a lot.

[None !niqjediPCA No.8856262]

>>8856246
>Each variety of cupcakes, as a whole (25 of the same type) gives us 25 possible combinations.

18 possible combinations.

>Each variety as a majority (24), with a single other cupcake of all of the other varieties, gives us an additional 24 possibilities.

additional 17 possible combinations.

>Each variety as a majority (23), with two cupcakes of a single variety, give us an additional 24 possibilities, etcetera...

additional 17 possible combinations.

[Anonymous No.8856296]

Assuming you can mix cupcakes and macrons in the same pack the solution is

[eqn]\left(\!\!{28 \choose 25}\!\!\right)={52 \choose 25} [/eqn]