K Permutations, 7 letter string, alphabet, no repeats
Number of 7 letter strings with no repeated letters can be formed from the english alphabet?
I get $$p(26,19)$$ so $$26! / 19!$$
but my answer sheet says it's:
$$ p(26,7) = 26\times25\times24\times23\times22\times21\times20$$
Can somebody confirm to me which ones correct please?
$\endgroup$ 13 Answers
$\begingroup$Both of your answers are equivalent, and therefore correct. By definition of factorial, $$\frac{26!}{19!}=\frac{26\times 25\times \dots \times 1}{19\times 18\times\dots\times 1}=26\times 25\times 24\times 23\times 22\times 21\times 20.$$
$\endgroup$ 7 $\begingroup$For unordered string $$\binom{26}{7}=\binom{26}{26-7}=\binom{26}{19}$$ for ordered 7-leter string $$\frac{26!}{(26-7)!}=\frac{26!}{19!}$$ for ordered 19-leter string $$\frac{26!}{(26-19)!}=\frac{26!}{7!}$$ as you ccan see $$\frac{26!}{19!}\neq\frac{26!}{7!}$$
$\endgroup$ $\begingroup$Presuming in your notation that you mean $p(26,7) = \binom{26}{7}$ and $p(26,19) = \binom{26}{19}$;
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
Hence;
$$\binom{26}{7} = \frac{26!}{7!19!}$$ $$\binom{26}{19} = \frac{26!}{19!7!}$$
Which are the same.
Note that $\frac{26!}{19!} \not = \frac{26!}{7!}$
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