For $(a + b + c + d)^{10}$, how many terms have coefficients that aren't divisible by $5$?

Question

$(a + b + c + d)^{10}$

how many terms have coefficients that aren't divisible by $5$?

I know that each of the following $a^{10}, b^{10}, c^{10}, d^{10} $, their coefficient is 1, which isn't divisible by 5. I also know that the sum of all the coefficients is :

$a=b=c=d=1$,

$(1+1+1+1)^{10} = 4^{10} = 1,048,576$

so if I subtract the coefficients I found so far, I'll still have 1,048,572, which still isn't divisible by 5, which means there's at least 1 more coefficient that's not divisible by 5. got stuck here. probably not even thinking in the right direction. please help :)

Answer 1

Hint: The coefficient of $a^{x}b^{y}c^{z}d^{w}$ in $(a+b+c+d)^{10}$ is $$\binom{10}{x}\binom{10-x}{y}\binom{10-x-y}{z}\binom{10-x-y-z}{w}.$$ And for $r = 0,\ldots,10,$ we have $$\binom{10}{r} = \dfrac{10!}{r!(10-r)!},$$ which is not divisible by $5$ if and only if $r = 0,5$, or $10$. (Why?)

Answer 2

Every term can be represented by a quadruple $(k,l,m,n)$ where the the entries are non-negative integers satisfying $k+l+m+n=10$.

The coefficient that corresponds with $(k,l,m,n)$ is $\frac{10!}{k!l!m!n!}$.

Observe that the numerator contains exactly two factors $5$, and conclude that the coefficient is not divisible by $5$ if also the denominator contains two factors $5$.

So we discern the following possibilities for not being divisible by 5:

• one of the entries of $(k,l,m,n)$ takes value $10$ (and consequently the others take value $0$)
• two of the entries of $(k,l,m,n)$ take value $5$ (and consequently the others take value $0$).

Evidently there are $4$ quadruples that satisfy the condition under the first bullet and there are $\binom42=6$ quadruples that satisfy the condition under the second bullet.

We conclude that $4+6=10$ terms have coefficients that are not divisible by $5$.