1hey subash if a+b+c......+k=n
Then n!/(a!b!c!...k!)=+ve integer
a,b,c......,k are positive integers such that a+b+c+....+k≤n, then can we say that n!/(a!b!c!.....k!) is also a positive integer??
solve with proof
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14 Answers
1wel it is definitely positive............but not sure whether this is an integer or not....................must solve w8.....
1lets 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
62when a+b+c+..k=n
n!/a!b!c!...k!
=nCa.n-aCb.n-a-bCc.....
each of these is an integer.. hence the above expression is integer..
341There 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 ep(k!) is given by
ep (k!) = [k/p] + [k/p2]+[k/p3]+.....
So ep (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/pr] ≥ [(a+b+c+d+...+k)/pr] ≥ [a/pr]+[b/pr]+...+[k/pr]
This immediately leads us to
ep(n!) ≥ ep(a!) + ep(b!)+...+ep(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 :)