combinatorics simple

wel it is definitely positive............but not sure whether this is an integer or not....................must solve w8.....

hw can sum of positive integers be negative telly????

hey telly tht is not necessary..............

oh ok!!!!!!!!
then
definiftely

n!/(a!b!c!...k!)=+ve integer

hey akand wat is

n! - (a+b+c+d.........+j)!=k! rite

lets take an example.....................
a=1 b=2 c=3 c=4 c=5 c=6........
n=1+2+3$+5+6+......=.....
so..........n!=sum huge number...
so d expression need not be an INTEGER

ok.............thnx bhaiyya

ok bhaiyya

what about the≤ part

There are two ways to go about this one.

First the claim is that the number is an integer

Algebraic Way: To prove that n!/ a!b!c!...k! is an integer, we will prove that

(1) Every prime that divides the denomintor also divides the numerator and

(2) Every such prime occurs with a higher power in the numerator than in the denominator

The first part is easy to prove as in fact a!, b!,..., k! are all divisors of n!

To prove the second part, we know that the maximal power of a prime that occurs in k! denoted by e_p(k!) is given by

e_p (k!) = [k/p] + [k/p²]+[k/p³]+.....

So e_p (n!) = [n/p] + [n/p²]+[n/p³]+.....

Now it is easily proved that [x+y] ≥ [x]+[y] which can be extended to

[x₁+x₂+....] ≥[x₁] + [x₂]+....

Hence we have

[n/p^r] ≥ [(a+b+c+d+...+k)/p^r] ≥ [a/p^r]+[b/p^r]+...+[k/p^r]

This immediately leads us to

e_p(n!) ≥ e_p(a!) + e_p(b!)+...+e_p(k!)

which means the power of p appearing in the numerator is not less than the power appearing in the denominator thus mkaing the given expression an integer

Combinatorial Argument

n!/(a!b!...k!) is just the number of permutations of n objects where a are identical in one way, b in another way and so on. So, if a+b+c+..+k ≤n, thats gotta be an integer :)