Math Puzzles

Here's a couple of math puzzles I came up with: Have Fun!


1.
If you have 36 different and separate objects, how many different combinations can you put them into? (the maximum).


2.
If you have 36 separate different objects, and each object came as either 'large' or 'small', how many different combinations could you come up with? (the maximum).


3.
There are 25 chairs, your mother is mean and you have to sit in them from chair to chair, in a correct predetermined order. If you make a wrong choice, you get slapped and made aware you made the wrong choice of chair. What is the maximum amount of times you can get slapped in the face?

just a guess:

36^36
72^72
25^25

??

Not the right answer, but somewhat on the right track...

The first one is 36! I think. I mean 36 factorial (36x35x34x33...x1).

the last would be 24+23+22+21+20+...

for the first chair, you would get slapped 24 times before you chose the right one.
Then 23 times for the next one
then 22 etc.

so total is 24 + 23 + 22 + ...
=24*25/2 = 300 times (ouch)

As for the first one, each object can be in or out of the selection
36 objects
=> 2^36 selections.

The first answer is approximately: 371,993,326,789,901,000,000,000,000,000,000,000,000,000
36*35*34... as nishantp said. However, this is assuming that each peice only has two sides that connect to the other peices. If it has say, 8, it would be that number ^8

With the same assumption, the seond one is approximately: 61,234,458,376,886,100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,0 00,000,000,000,000,000,000,000,000,000

Which, if you're wondering, is more than the number of atoms in the known universe!!

I don't know about the chair thing, but waht sql_lall said makes sense to me.:D

36! is correct for #1.

300 is the correct answer for #3.

I'll post the answer (and proof) for #2 after more people get a chance at it.

Wouldn't #2 be:

72!

That's what I thoguht. That's what the really huge number in my first post is.(The second really huge number)

#2 = (36!)^2 ???

Quote:

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Originally posted by Muddy
#2 = (36!)^2 ???

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It might be. Thats a really huge number. About 1.384 x 10⁸³.

"If you have 36 different and separate objects, how many different combinations can you put them into? (the maximum)."

Just wondering, by "different combinations" do you mean:
1) A subset of these 36, where order is irrelevant
i.e. {A,B,C} is the same as {A,C,B}, and {A,B,D,C} is also allowed
In this case, it's 2^36

2) A subset where order is important
i.e. {A,C,D,B} is different to {D,B,C,A} etc.
in this case, i'm not exactly sure how many, but i it would be between 36! and 2^36 * (36!)
Well, the exact value is the Sum{i=0 to 36} [³⁶ P _i]

3) A permutation of the 36, i.e. how many ways are there to order these 36, which would be 36!

4) taken literally, ³⁶ C ₃₆ = 1

Quote:

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Originally posted by Muddy
#2 = (36!)^2 ???

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That would be my guess, if 72! is wrong.

;) #2 was a 'trick' question.

The answer is 72!

i don't see how:

72! would only be true if there were 72 objects. However, you said that each of the 36 distinct objects came in EITHER large OR small, so could only be one of them

=> For each object, you have either large or small (2^36 ways)
then you have 36! ways of ordering these.

=> 2^36 * 36! ways
(N.B. this is less that 72!)

Either can imply one and the other.

Such as a person wearing rings on both hands. You could say she's wearing rings on either hand.

http://dictionary.reference.com/search?q=either

Enough of a usage note to make you not really care. :p

Ahh yes, a dictionary, the mathematicians enemy.

However, if we are being semantic, i think you will find the question states 'either type_a OR type_b'

This is very obviously an either-or construction, which is closest to the Conjunction definition of either, where you have two options and can choose one.

I think you can only use 'either' as inclusive for certain objects (like hands), but for everything else you use BOTH. (i.e. the objects came in BOTH type_a and type_b)

Anyway, i guess neither of us really care :rolleyes:

I'll just say the answer is 2^36 x 36!, and close dictionary.reference.com as quickly as possible.

I am not sure that I understand the question or the answers at this thread.

Combinations is a mathematical term which implies that order does not matter. The factorial function is used to calculate permutations, for which order is significant.

For example, one might ask how many combinations of 3 objects can be made from a set of 7 distinct objects.

Answer: 35 = 7! / 3!4!

How many permutations? 210 = 7! / 4! (or 7*6*5)

Combinations of 36 objects seems ambiguous.

If the questions asks how many ways can 36 distinct objects be permuted or arranged, the answer is 36!

If it asks how many combinations of 36 distinct objects can be made from a set of 36 objects, the answer is only one.

If the question is how many combinations of any number of the 36 objects can be made, the answer is a lot more than one.

There is one combination using no objects. There are 36 combinations using any one of the 36 objects. There are 630 combinations using pairs of the 36 objects. Et cetera for groups of 3, 4, 5 . . .

The answer to the latter interpetation is 2³⁶, which is a lot less than 36!

The answer to the second question depends on interpretation.

1, 2³⁶, 2⁷², 36!, or 72!