today I've encountered a problem like the following:

One of the terms in the open form of $(3x^2+2x+y+4z)^{10}$ is randomly chosen, what is the probability that the chosen term contains $x^7$?

My Attempts

I've reduced the question to two pieces, -calculating the number of terms with $x^7$ and the number of all terms.

I've opened the brackets using multinomials: $$\sum_{k_1,k_2,k_3,k_4=1\\k_1+k_2+k_3+k_4=10}^{10} \dbinom{10}{k_1,k_2,k_3,k_4} (3x^2)^{k_1}\cdot(2x)^{k_2}\cdot y^{k_3}\cdot(4z)^{k_4}$$ To calculate the number of terms with $x^7$ I've used that $2k_1+k_2=7$ easily I got $(0,7),(1,5),(2,3),(3,1)$ as the number of solutions for this, thus $4$ terms with $x^7$. However when it got to calculating the number of all terms It got a little more complicated, As an initial thought $x^2$ allows us to get $x^{20}$ for the max degree and $x^{10}$ for the minimum, I assumed that they can take all the values between firstly and what I got was $\{10,11,12,13,14,15,16,17,18,19,20\}$ thus I said the answer might be $\dfrac{4}{11}$. Though I think that my solution has technical errors, and the answer I gave isn't in the options.

What are your suggestions?

  • $\begingroup$ By "open form" do you mean that all coefficients are $1$? That is, is the open form of $(x+1)^2$ given by $x^2 + 2x + 1$ or $x^2 + x + x + 1$? $\endgroup$ Dec 18, 2017 at 18:54
  • $\begingroup$ I think that it is more like $x^2+2x+1$. But of course the question might have meant otherwise, but the options a,b,c,d,e didn't look much like the second one in which the coefficients were $1$. $\endgroup$ Dec 18, 2017 at 19:00

1 Answer 1


Just enumerate the possibilities:

  1. $k_1=0, k_2=7$: There are 4 possibilities for $(k_2, k_3)$ (in order that $k_1+k_2+k_3+k_4=10$), i.e. $(0, 3), (1, 2), (2, 1), (3, 0)$.

  2. $k_1=1, k_2=5$: There are 5 possibilities


The above enumeration generates 22 terms. Since there are $\binom{13}{3} = 286$ terms in total, we get $22/286 = 11/143 = 1/13$

  • $\begingroup$ How did we end up with $4^{10}$ terms at total? Don't we maximally have $k_1+k_2+k_3+k_4=10\implies \dbinom{10+3}{3}$ terms? (And that even doesn't count in the situations in which we have two different degrees of $x$) $\endgroup$ Dec 18, 2017 at 19:17
  • $\begingroup$ Your logical way is best. For those of us on the lazy side: With Mathematica one could execute poly = Expand[(3 x^2 + 2 x + y + 4 z)^10]; Length[Coefficient[poly, x^7] /. Plus -> List]/Length[poly /. Plus -> List] $\endgroup$
    – JimB
    Dec 19, 2017 at 1:50

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