Probability help (1 Viewer)
- Thread starter apollo1
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D94
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(a): you have 5 letters, so to have 2 letters be A, it's 5C2. And then the other three spaces can have a maximum of 2 letters each (ie. B and C), therefore 2*2*2 = 2^3. So you get 5C2 * 2^3
(b):
Since we only want even and zero cases, consider the binomial expansion of the following two cases. Since the second case alternates signs (when n is even), we can add the two expansions together to get only the even and zero cases.
^{2n} \\ \\= \binom{2n}{0}2^{2n} + \binom{2n}{1}2^{2n-1} + \binom{2n}{2}2^{2n-2} + \binom{2n}{3}2^{2n-3} + \binom{2n}{4}2^{2n-4} + \cdots + \binom{2n}{2n} \\\\ \\(1-2)^{2n} \\ \\= \binom{2n}{0}2^{2n} - \binom{2n}{1}2^{2n-1} + \binom{2n}{2}2^{2n-2} - \binom{2n}{3}2^{2n-3} + \binom{2n}{4}2^{2n-4} - \cdots + \binom{2n}{2n} \\\\ \\(1+2)^{2n} + (1-2)^{2n}=2\left ( \binom{2n}{0}2^{2n} + \binom{2n}{2}2^{2n-2} + \binom{2n}{4}2^{2n-4} + \cdots + \binom{2n}{2n} \right ) \\ \\\frac{(3)^{2n} + (-1)^{2n}}{2}\ \ = \2^{2n} + \binom{2n}{2}2^{2n-2} + \binom{2n}{4}2^{2n-4} + \cdots + 1 \\=\frac{1}{2} (3^{2n} + 1)\ when\ n \ is\ even\ or\ zero.)
(b):
Since we only want even and zero cases, consider the binomial expansion of the following two cases. Since the second case alternates signs (when n is even), we can add the two expansions together to get only the even and zero cases.