Binomial coefficients and multinomial coefficients

Win_odd Dhamnekar

Win_odd Dhamnekar

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We know that \(\displaystyle \binom{n}{0} + \binom{n}{1} + \binom {n}{2} + ... + \binom {n}{n}= 2^n \rightarrow (1)\)

Now, if I want to have R.H.S of (1) as 3n, how should I adjust L. H. S. of (1)?
 
Just multiply the left hand side by (2/3)^n. Your question is basically what do you multiply 2^n by to get 3^n.
 
Dhamnekar Winod said:
We know that \(\displaystyle \binom{n}{0} + \binom{n}{1} + \binom {n}{2} + ... + \binom {n}{n}= 2^n \rightarrow (1)\)

Now, if I want to have R.H.S of (1) as 3n, how should I adjust L. H. S. of (1)?
Click to expand...
It's easier to see using the binomial theorem.
[math](x+y)^n=\sum_{k=0}^{n} {n \choose k}x^{k}y^{n-k}[/math]For [imath]x=1,y=1[/imath], we have
[math]\tag*{Same as the your equation (1)}(1+1)^n=2^n=\sum_{k=0}^{n}{n \choose k}[/math]Now, let [imath]x=2,y=1[/imath], you shall have your answer.
 
Last edited:
BBB,
While I am sure that your answer is what was expected, I do not think that you adjusted the lhs. You completely changed it!
 
Steven G said:
BBB,
While I am sure that your answer is what was expected, I do not think that you adjusted the lhs. You completely changed it!
Click to expand...
I agree if that's, in fact, the exact wording of the author. There's a high probability that "adjust" is the OP's word choice or translated, and not the author's. One can only guess.
 
The left-hand side could be rewritten as:

\(\displaystyle \bigg[1 \binom{n}{0} + \dfrac12\binom{n}{1} +\dfrac14 \binom {n}{2} + ... + \dfrac{1}{2^n}\binom {n}{n}\bigg] \bigg[\binom{n}{0} + \binom{n}{1} + \binom {n}{2} + \binom {n}{3} + ... + \binom {n}{n}\bigg]\)


Example:

\(\displaystyle 3^2, \ \) where n = 2 in the formula, would be

[1 + 1 + 0.25][1 + 2 + 1] =

[2.25][4] =

9

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You may be looking for just one sum expression.
 
Answer to this question for n=5 :\(\displaystyle \binom{5}{5,0,0} + \binom{5}{0,5,0} + \binom{5}{0,0,5} + \binom{5}{4,1,0} +\binom{5}{4,0,1} + \binom{5}{1,4,0} + \binom{5}{0,4,1} + \binom{5}{3,2,0} + \binom{5}{3,0,2} + \binom{5}{0,3,2} + \binom{5}{2,3,0} + \binom{5}{3,1,1} + \binom{5}{1,3,1} + \binom{5}{1,1,3} + \binom{5}{2,2,1}+ \binom{5}{2,1,2} + \binom{5}{1,2,2} + \binom{5}{1,0,4} + \binom{5}{0,1,4} + \binom{5}{2,0,3} + \binom{5}{0,2,3}= 1+1 +1 +5 + 5 + 5 + 5+ 10 +10 +10 +10 + 20 + 20 + 20 + 30 + 30 + 30 + 5 + 5 + 10 + 10 =243 = 3^5\)
 
So, we can form multinomial combination sum for (a + b + c)n= 3n, (a + b + c + d)n = 4n, (a + . . . + k)n= kn
 
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