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Question(a) Find the number of 5-letter passwords with no repeated lowercase letters from 26 alphabet letters. (b) A company has 33 salespeople. Find the number of possible rankings of the top 3 salespeople.
(a) (265)\binom{26}{5} (b) (333)\binom{33}{3}

Studdy Solution

STEP 1

Assumptions1. For the computer password, no letter may be repeated and only the lower case of a letter may be used. There are26 letters in the alphabet. . For the salespeople ranking, there are33 salespeople in total and we need to find the number of ways to rank the top3.

STEP 2

For part (a), we need to find the number of ways to choose5 letters from26 without repetition. This is a permutation problem, and the formula for permutations is(n,r)=n!(n−r)!(n, r) = \frac{n!}{(n-r)!}where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

STEP 3

Now, plug in the given values for n and r to calculate the number of possible passwords.
(26,5)=26!(26−5)!(26,5) = \frac{26!}{(26-5)!}

STEP 4

Calculate the number of possible passwords.
(26,)=26!21!(26,) = \frac{26!}{21!}

STEP 5

implify the factorial expression to get the number of possible passwords.
(26,5)=26×25×24×23×22=7,893,600(26,5) =26 \times25 \times24 \times23 \times22 =7,893,600

STEP 6

For part (b), we need to find the number of ways to rank the top3 salespeople out of33. This is also a permutation problem, and we can use the same formula as in part (a).

STEP 7

Now, plug in the given values for n and r to calculate the number of possible rankings.
(33,3)=33!(33−3)!(33,3) = \frac{33!}{(33-3)!}

STEP 8

Calculate the number of possible rankings.
(33,3)=33!30!(33,3) = \frac{33!}{30!}

STEP 9

implify the factorial expression to get the number of possible rankings.
(33,3)=33×32×31=32,856(33,3) =33 \times32 \times31 =32,856So, there are7,893,600 possible5-letter passwords and32,856 possible rankings of the top3 salespeople.

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