Answer to Question #192060 in Discrete Mathematics for ANJU JAYACHANDRAN

Question #192060

Use generating function to prove the identity

n

k=o



r

k

 s

n−k



=



r +s

n





1
Expert's answer
2021-05-13T01:18:41-0400

"\\sum_{k=0}^n\\binom rk\\binom s{n-k}=\\binom {r+s} {n }"


we know, "(1+x)^{r+s}=(1+x)^r(1+x)^s"


"\\sum_{k=0}^{r+s}\\binom {r+s}{n}x^n=\\sum_{j=0}^r \\binom{r}{j}x^j+\\sum_{a=o}^s\\binom{s}{a}x^a"


Now we expand each of the summation on RHS upto "x^2" -


"=[\\binom r0+\\binom r1x+\\binom r2x^2+....][\\binom s0+\\binom s1x+\\binom s2 x^2+...........]\n\n\n\\\\[9pt]=[\\binom r0\\binom s0]+[\\binom r0\\binom s1+[\\binom s0\\binom r1]x+[\\binom r1\\binom s1+\\binom r0\\binom s2+\\binom r2\\binom s0]x^2......."


Thus By inspecting a clear Pattern is evident and the n^{th} Power of ''x'' is of the form -


"\\sum _{k=0}^n \\binom rk \\binom s{n-k}x^k"


and This is equated to "\\binom {n+m}{k} x^k."


Hence, "\\sum_{k=0}^n\\binom rk\\binom s{n-k}=\\binom {r+s} {n }"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS