EVALUATE Σi≠j 21Ci*21Cj where 0 ≤ i ≤10 , 0 ≤ j ≤ 10.
62Hint:
this is equal to
1/2 { Σi≠j 21Ci*21Cj} where 0 ≤ i ≤21 , 0 ≤ j ≤ 21
which is equal to
1/2 { Σi [Σj 21Ci*21Cj] - Σi=j 21Ci*21Cj}
1actually i am not able to find mistake in the following solution. this is what i did to solve the question .
WE NEED TO FIND
21C0 {21C1+21C2+21C3......21C10} +21C1 {21C0+21C2+21C3.....+21C10} +21C2{21C0 +21C1+21C3+.........+21C10}................+21C10{21C0+21C1+21C2+..............21C9}
AM I RIGHT??
SO WE JUST NEED TO FIND OUT
{21C0+21C1+21C2+........21C10}2 -{21C02+21C12+21C32+......+21C102}
I GOT THE ANSWER AS 240-{42C21}/2 ..BUT THE ANSWER IS 1/2{240-(42C21)/2}
62yes this is perfect looking..
Or may be even I am out of my mind right now :)
But somehow I think what you have done is correct!
341Your answer is right. If you have practice in using summation symbols you can write it more succintly and with more clarity.
The problem asks us to find \sum_{i \ne j} \binom {21}{i} \binom {21} {j}
Now we write
\sum_{i \ne j} \binom {21}{i} \binom {21} {j} = \sum_{i=0}^{10} \sum_{j=0}^{10} \binom {21}{i} \binom {21} {j} - \sum_{i=0}^{10}\binom {21}{i}^2 \\ \\ = \sum_{i=0}^{10} \binom {21}{i} 2^{20} - \frac{1}{2} \binom {42}{21} \\ \\ = 2^{40} - \frac{1}{2} \binom {42}{21}
21
Sir can u pl. explain how did u get the purpled STEPS.... [7]
341\sum_{i=0}^{10} \sum_{j=0}^{10} \binom {21}{i} \binom {21} {j} = \sum_{i=0}^{10} \binom {21}{i} \sum_{j=0}^{10}\binom {21} {j}
Now
\sum_{j=0}^{10}\binom {21} {j} = \frac{1}{2} \sum_{j=0}^{21} \binom {21} {j} = \frac{1}{2} \times 2^{21} = 2^{20}
And
\sum_{i=0}^{10}\binom {21} {i} \sum_{j=0}^{10}\binom {21} {j} = \sum_{i=0}^{10}\binom {21} {i} 2^{20} = 2^{20} \sum_{i=0}^{10}\binom {21} {i} = 2^{20} \times 2^{20} = 2^{40}
21OH K!!!!
I NEVER REALIASED, da fact wich u wrote in ur step1 of post #8
THNK YOU SIR[1]