Simplify this expression...

(x-a)(x-b)(x-c)....(x-z)

(x-a)(x-b)(x-c)....(x-n) =x^N.........(abc...n)

HAHAHA



are the numbers a, b etc random, or to they have some relationship

Since the indicated product clearly includes "(x-x)", the product is 0! :D

The expansion is:
x^n+(a+b+c...z)x^(n-1)+(ab+bc+ac...yz)x^(n-2)+...+(abc...z)
Any particular reason why?

combinations and permutations.

Figure out how many ways you can permute a set of m symbols, of n elements, then sum over all sets. For example. given the set of all elements in the alphabet, i.e. m=26, how many configurations exist of 2 elements such that each configuration is commutative and may only appear once. Answer (26*25)/2.

What about all sets containg 1-26 elements

answer (26 C 0)+(26 C 1) + (26 C 2) + ...............( 26 C 26)

"What about all sets containg 1-26 elements

answer (26 C 0)+(26 C 1) + (26 C 2) + ...............( 26 C 26)"
Which is, of course, 2^26.

(By the way, since you start with 26C0 instead of 26C1, it is the number of subsets of a set with 26 elements, each subset containing 0 to 26 elements, not 1-26.)

Originally posted by HallsofIvy
Since the indicated product clearly includes "(x-x)", the product is 0! :D

Since 0! = 1, then I understand that you mean that the product = 1. :D

I don't understand how.
I believe that if the product includes "(x-x)", the product should be 0.
However, as far as I remember the English alphabet is:
a b c d e ... m n o p ... w x y z.
Since the product stops at n which is before x in the alphabet, "(x-x)" is not included in the product which is consequently not equal to 0. :D

Originally posted by 1100f
Since 0! = 1, then I understand that you mean that the product = 1. :D

I don't understand how.
I believe that if the product includes "(x-x)", the product should be 0.
yes

However, as far as I remember the English alphabet is:
a b c d e ... m n o p ... w x y z.
Since the product stops at n which is before x in the alphabet, "(x-x)" is not included in the product which is consequently not equal to 0. :D

the product stops at z. don t mistake the original post in this thread by speeding electron for ryans post, which was the second post in this thread.

speeding electron wrote a product that went to z, but ryans missed the trick, and tried to do the problem the hard way, and therefore failed to get the correct answer.

ryans only expaned n terms, when he should have either expanded 26 terms explicitly, or simply noticed that (x-x) is one of the terms.

so make sure you are reading the correct version of the question.

Originally posted by lethe
.

so make sure you are reading the correct version of the question.

You are right. I am left (with my mistake).

Which leaves me in the middle.

How silly of me.
Here was I, assuming (x-a)(x-b)...(x-z) indicated an expansion of an arbitrary number of terms.

If we're having fun...
A sodium ion walks into a bar, and asks the barman for a packet of crisps. The barman says "Are you sure?" to which the ion replies "Yes, I'm positive.":D

Originally posted by geodesic
and asks the barman for a packet of crisps.

what are crisps?

What you would call chips.

What do you call a joke involving Cobalt, Radon and Yttrium?

CoRnY.