i'm opening this for all stupid math questions

and here's mine

how would one calculate the number of potential combinations, ie, 5 spaces, 26 letter alphabet, what are the number of possible letter combinations.

1.2 yes that is if you are useing the old sliding scale approch if not it is 3.2, although many would argue witht he tecnologies these days you could be wrong in saying that is ethier and in fact the answer to this question is indeed 0(zero)
then again that is only three years of research. When i asked my lecturer this question they said
"the question with many answers is like is as good as a question with no answers"
.........

eisbaer:

<i>how would one calculate the number of potential combinations, ie, 5 spaces, 26 letter alphabet, what are the number of possible letter combinations.</i>

In the first space, you can put any of 26 letters.
In the second space, you can put any of 26 letters.
etc.

So, the number of combinations is 26⁵.

So, the number of combinations is 26^5.
Assuming you can reuse every letter each space.

If not...

If each letter may only be used in 1 place (ie, {a,a,b,c,d} is not a valid solution because it contains 2 "a" letters), then the number of possibile combinations is 7,893,600 or 26x25x24x23x22.

If the order of letters matters (ie, {a,b,c,d,e} is the same thing as {e,d,c,b,a} or {a,c,d,e,b} or any combination that contains each of those letters), then the number of possibile combinations drops to 65,780.

Here's a stupid one:

what comes next ?

4, 5, 5, 7, 14, 17, 51, 55, __

380
220
99
256

I'm ashamed to say I have no clue :eek:

220.
Pay attention that:
5/4=1.25
*5/5=1
7/5=1.4
*14/7=2
17/14=1.21
*51/17=3
55/51=1.08

look at the ones with "*". The next one should be 4. Thus, 55*4=220.

BSG's answer is correct, but just to add a little to the explanation-

4, 5, 5, 7, 14, 17, 51, 55, ___

4+1 = 5
5*1 = 5
5+2 = 7
7*2 = 14
14+3 = 17
17*3 = 51
51+4 = 55
55*4 = 220

Orbie,

If each letter may only be used in 1 place (ie, {a,a,b,c,d} is not a valid solution because it contains 2 "a" letters), then the number of possibile combinations is 7,893,600 or 26x25x24x23x22.

If the order of letters matters (ie, {a,b,c,d,e} is the same thing as {e,d,c,b,a} or {a,c,d,e,b} or any combination that contains each of those letters), then the number of possibile combinations drops to 65,780.

I don't understand the reasoning of this. If the order of the objects matters, then the number of possible outcomes goes down?

I think you're confusing permutations (<i>ⁿP_r</i> - the number of <i>ordered</i> outcomes by choosing <i>r</i> objects out of a choice of <i>n</i>) and binomial coefficients (<i>ⁿC_r</i> - choosing <i>r</i> objects out of a choice of <i>n</i> in an <i>unordered</i> fashion).

The values you state would indicate that this is the case.

_________________________

Eisbar,

To recap the things said here:

If the letters can be used more than once, then it is the number of letters raised to the power of the number of spaces available. In this case 26⁵ or 11,881,376.

If the letters can only be used once, and the ordering of the letters is important, then the number of permutations is the factorial of the number of letters divided by the factorial of the difference of the number of letters and the number of spaces:

<i>n!</i> / (<i>n-r</i>)<i>!</i>

Where <i>n</i> is the number of letters and <i>r</i> the number of spaces. This has a value of 7,893,600.

If the order is unimportant, then it is number of permutations divided by the factorial of the number of spaces. Algebraicly:

<i>n!</i> / (<i>n-r</i>)<i>!r!</i>

The value in this case is 65,780.