3 years ago

In how many ways can a set of five letters be selected from the english alphabet

1 answer:

3 0

Without replacement 26*25*24*23*22

with replacement 26^5

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Ira Lisetskai [31]

Y > 20

Hope this helps! Message back for an explanation:)

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Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$

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AysviL [449]

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prisoha [69]

Answer:

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Step-by-step explanation:

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