Three people from a group of twelve are needed for a committee. In how many different ways can a committee be formed?

 

Answer: The twelve folk are to be labeled as follows: 3 as “on the committee” and 9 as “lucky.” The answer must be \(\dfrac{12!}{3!9!}=220\).

 

COMMENT: Notice that we were sure to assign an appropriate label to each and every person (or object) in the problem.  This fits the “self check” we described earlier: the sum of the labels represented in the denominator should match the count of objects represented in the numerator.

 

Fifteen horses run a race. How many possibilities are there for first, second, and third place?

 

Answer: One horse will be labeled “first,” one will be labeled “second,” one “third,” and twelve will be labeled “losers.” The answer must be: \(\dfrac{15!}{1!1!1!12!} = 15 \times 14 \times 13 = 2730\) .

 

A “feel good” running race has 20 participants. Three will be deemed equal “first place winners,” five will be deemed “equal second place winners,” and the rest will be deemed “equal third place winners.” How many different outcomes can occur?

 

Answer: Easy! \(\dfrac{20!}{3!5!12!}\).

 

 

From an office of 20 people, two committees are needed. The first committee shall have 7 members, one of which shall be the chair and one the treasurer. The second committee shall have 8 members. This committee will have 3 co-chairs and 2 co-secretaries and 1 treasurer. In how many ways can this be done? (Assume no one is to serve multiple roles.)

 

Answer: Keep track of the labels. Here they are:

            1 person will be labeled “chair of first committee”

            1 person will be labeled “treasure of first committee”

            5 people will be labeled “ordinary members of first committee”

            3 people will be labeled “co-chairs of second committee”

            2 people will be labeled “co-secretaries of second committee”

            1 person will be labeled “treasurer of second committee”

            2 people will be labeled “ordinary members of the second committee”

            5 people will be labeled “lucky,” they are on neither committee.

 

The total number of possibilities is thus: \(\dfrac{20!}{1!1!5!3!2!1!2!5!}\) . Piece of cake!

 

 

Suppose 5 people are to be chosen from 12 and the order in which folk are chosen is not important. How many ways can this be done?

 

Answer:  5 people will be labeled “chosen” and 7 “not chosen.” There are \(\dfrac{12!}{5!7!}\)  ways to accomplish this task.

 

 

Suppose 5 people are to be chosen from 12 for a team and the order in which they are chosen is considered important. In how many ways can this be done?

 

Answer: We have:

                                    1 person labeled “first”

                                    1 person labeled “second”

                                    1 person labeled “third”

                                    1 person labeled “fourth”

                                    1 person labeled “fifth”

                                    7 people labeled “not chosen”

 (Again … make sure everyone has a label!)   This can be done \(\dfrac{12!}{1!1!1!1!1!7!}\)  ways.

 

 

 

That’s it. This is basically the end of the story! (The fourth step in this course is just learning how to combine the multiplication principle and the labeling principle together in stages.)

 

It’s all easy and straightforward!

 

SOME PRACTICE: 

Remember, all solutions appear in the COMPANION GIUDE to this Permutations and Combinations course.

 

Exercise 17:a)    In how many different ways can one arrange five As and five Bs.b)    A coin is tossed 10 times. In how many different ways could exactly five heads appear?

 

 

Exercise 18:  The word BOOKKEEPING is the only word in the English language (ignoring its variants: bookkeeper, etc.) with three consecutive double letters. In how many ways can one arrange the letters of this word?

 

 

Exercise 19:a)    A mathematics department has 10 members. Four members are to be selected for a committee. In how many different ways can this be done?b)    A physics department has 10 members and a committee of four is needed. In that committee, one person is to be selected as “chair.” In how many different ways can one form a committee of four with one chair?c)     An arts department has 10 members and a committee of four is needed. This committee requires two co-chairs. In how many different ways can one form a committee of four with two co-chairs? d)    An English department has 10 members and two committees are needed: One with four members with two co-chairs and one with three members and a single chair. In how many different ways can this be done?

 

 

Exercise 20: Three people from 10 will be asked to sit on a bench: one on the left end, one in the middle, and one on the right end. In how many different ways can this be done?

 

 

Exercise 21:   Five pink marbles, two red marbles, and three rose marbles are to be arranged in a row. If marbles of the same color are identical, in how many different ways can these marbles be arranged?

 

 

Exercise 22: a)    Hats are to be distributed to 20 people at a party. Five hats are red, five hats are blue, and 10 hats are purple. In how many different ways can this be done? (Assume the people are mingling and moving about.)b)    CHALLENGE: If the 20 people are clones and cannot be distinguished, in how many essentially different ways can these hats be distributed?

 

Exercise 23: In how many ways can one arrange the letters of NOODLEDOODLE if the arrangement must begin with an L and end with an E?

 

 

Exercise 24: a) In how many ways can the letters ABCDEFGH be arranged?a) In how many ways can the letters ABCDEFGH be arranged with letter G appearing somewhere to the left of letter D?b) In how many ways can the letters ABCDEFGH be arranged with the letters F and H adjacent?

 

 

 

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