Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We know that $(1+x)^2$ has $3$ distinct terms because $(1+x)^n$ has $n+1$ terms going by the popular expansion starting from ${}_nC_0$ to ${}_nC_n$.

How do we find total number of distinct terms in expressions like $(a+b+c+f)^{40}$ and what's the generalized result?

share|improve this question
1  
You might find this helpful Number of Terms in a Polynomial Expansion. Regards –  Amzoti Jan 21 '13 at 17:46
1  
en.wikipedia.org/wiki/… –  lab bhattacharjee Jan 21 '13 at 17:46
    
SO the answer is $39C3$? –  Bazinga Jan 21 '13 at 17:55

1 Answer 1

The Multinomial Theorem states that $$\left(\displaystyle\sum_{i=1}^k x_i\right)^n = \displaystyle\sum_{n_1 + \dots + n_k = n} \binom{n}{n_1, \dots, n_k}x_1^{n_1}\dots x_k^{n_k}$$ where $$\binom{n}{n_1, \dots, n_k} = \frac{n!}{n_1!\dots n_k!}.$$ So the number of terms in the expansion is equal to the number of non-negative solutions to the equation $n_1 + \dots + n_k = n$, which is $\displaystyle\binom{n+k-1}{n}$ as is proved here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.