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Chapter 6: Problem 25

How many three-letter (unordered) sets are possible that use the letters \(\mathrm{q}, \mathrm{u}, \mathrm{a}, \mathrm{k}, \mathrm{e}, \mathrm{s}\) at most once each?

Short Answer

Expert verified
There are 20 three-letter unordered sets possible using the letters q, u, a, k, e, s at most once each. We found this by applying the combination formula \(C(6, 3) = \frac{6!}{3!(6-3)!}\) and calculating the required factorials.

Step by step solution

01

Understand the combination formula

The combination formula helps us calculate the number of possible combinations (unordered sets) of a larger set (in this case, the set of six letters). The combination formula is given by: \(C(n, r) = \frac{n!}{r!(n-r)!}\) where C(n, r) is the number of combinations of n things taken r at a time, n! is the factorial of n, and r! is the factorial of r.
02

Identify n and r values

In our problem, the set of letters (q, u, a, k, e, s) has a total of 6 elements, so n = 6. We need to create three-letter combinations from this set, so r = 3.
03

Apply the combination formula

We can now apply the combination formula to our problem, using n = 6 and r = 3: \(C(6, 3) = \frac{6!}{3!(6-3)!}\)
04

Calculate the factorials

We need to calculate the factorial values for 6!, 3!, and (6-3)!: 6! = 6 脳 5 脳 4 脳 3 脳 2 脳 1 = 720 3! = 3 脳 2 脳 1 = 6 (6-3)! = 3! (since it evaluates to the same as 3!) = 6
05

Plug in the factorial values

Now, substitute the calculated factorials back into the combination formula: \(C(6, 3) = \frac{720}{6 \times 6}\)
06

Calculate the final result

We now calculate the total number of three-letter unordered sets: \(C(6, 3) = \frac{720}{36} = 20\) So, there are 20 three-letter unordered sets possible using the given letters at most once each.

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Most popular questions from this chapter

Let \(A=\\{\) June, Janet, Jill, Justin, Jeffrey, Jello\\}, \(B=\\{\) Janet, Jello, Justin\\}, and \(C=\\{\) Sally, Solly, Molly, Jolly, Jello\\}. Find each set. $$ (A \cap B) \cap C $$

Let \(S\) be the set of outcomes when two distinguishable dice are rolled, let \(E\) be the subset of outcomes in which at least one die shows an even number, and let \(F\) be the subset of outcomes in which at least one die shows an odd number. List the elements in each subset given. $$ F^{\prime} $$

How many different four-letter sequences can be formed from the letters a, a, a, b? HINT [See Example 3.]

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Is the set of outcomes when two indistinguishable dice are rolled (Example 1) a Cartesian product of two sets? If so, which two sets; if not, why not?
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