Discrete math
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Please answer ASAP! Discrete math
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Please answer ASAP! Discrete math
Please answer ASAP!
Discrete math
2. Use a combinatorial argument to prove that n Σ(0) - 2k =3" (2) =0 Solution: Let us take the set of all words of n letters from the alphabet A = {a,b,c}. We know there are 3such words. That is the right hand side of (2). Now, we partition this set into classes Ck, k = 0,1,2,..., n. For each such k lets put into Cx the words which contain exactly k a's. If k = 0, for instance then Co consists of all the n-letter words that are from the alphabet B = {b,c}. That means, we have Col = 2” which is the last term in the left hand side of (2). If say k = 3, we need to pick the 3 positions for a's and then complete the rest of the word with letters from B. This implies 161 = () 2-3 72 n n In general, we have in a similar counting |Cx) = (2) 2n-k. Hece, we must have 31 E 2 n -Σ(1)e--Σ(...)=Σ2 (7) 3 () Ž>) 120k 2nk= n- =0 ko 1 and so (2) is established.
Discrete math
2. Use a combinatorial argument to prove that n Σ(0) - 2k =3" (2) =0 Solution: Let us take the set of all words of n letters from the alphabet A = {a,b,c}. We know there are 3such words. That is the right hand side of (2). Now, we partition this set into classes Ck, k = 0,1,2,..., n. For each such k lets put into Cx the words which contain exactly k a's. If k = 0, for instance then Co consists of all the n-letter words that are from the alphabet B = {b,c}. That means, we have Col = 2” which is the last term in the left hand side of (2). If say k = 3, we need to pick the 3 positions for a's and then complete the rest of the word with letters from B. This implies 161 = () 2-3 72 n n In general, we have in a similar counting |Cx) = (2) 2n-k. Hece, we must have 31 E 2 n -Σ(1)e--Σ(...)=Σ2 (7) 3 () Ž>) 120k 2nk= n- =0 ko 1 and so (2) is established.
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