Suppose we want to choose 6 letters, without replacement, from 12 distinct letters.(If necessary, consult a list of formulas.)(a) How many ways can this be done, if the order of the choices is taken into consideration?(b) How many ways can this be done, if the order of the choices is not taken into consideration?
Question
Suppose we want to choose 6 letters, without replacement, from 12 distinct letters.(If necessary, consult a list of formulas.)(a) How many ways can this be done, if the order of the choices is taken into consideration?(b) How many ways can this be done, if the order of the choices is not taken into consideration?
Solution
(a) If the order of the choices is taken into consideration, we use the permutation formula. The number of ways to choose 6 letters from 12, without replacement and where order matters, is given by the formula for permutations of n items taken r at a time, denoted as P(n, r).
P(n, r) = n! / (n - r)!
Here, n is the total number of items to choose from (12 letters), and r is the number of items chosen (6 letters). The exclamation point denotes a factorial, which means to multiply that number by all positive integers less than it.
P(12, 6) = 12! / (12 - 6)!
= 12! / 6!
= 665,280
So, there are 665,280 ways to choose 6 letters from 12 when order matters.
(b) If the order of the choices is not taken into consideration, we use the combination formula. The number of ways to choose 6 letters from 12, without replacement and where order does not matter, is given by the formula for combinations of n items taken r at a time, denoted as C(n, r).
C(n, r) = n! / [r!(n - r)!]
Here, n is the total number of items to choose from (12 letters), and r is the number of items chosen (6 letters).
C(12, 6) = 12! / [6!(12 - 6)!]
= 12! / [6! * 6!]
= 924
So, there are 924 ways to choose 6 letters from 12 when order does not matter.
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