10.6: Combinations- Involving Several Sets

Learning Objectives

In this section, you will learn to

1. count the number of items selected from more than one set
2. count the number of items selected when there are restrictions on the selections

So far, we have solved the basic combination problem of \(\mathrm{r}\) objects chosen from n different objects. Now we will consider certain variations of this problem.

Example \(\PageIndex{1}\)

How many five-person committees consisting of 2 men and three women can be chosen from a total of 4 men and four women?

Solution

We list four men and four women as follows:

\[M_1M_2M_3M_4W_1W_2W_3W_4 \nonumber \]

Since we want 5-person committees consisting of 2 men and three women, we'll first form all possible two-person committees and all possible three-woman committees. Clearly there are 4C2 = 6 two-man committees, and 4C3 = 4 three-woman committees, we list them as follows:

2-Man Committees	3-Woman Committees
\[\begin{array}{l} \mathrm{M}_{1} \mathrm{M}_{2} \\ \mathrm{M}_{1} \mathrm{M}_{3} \\ \mathrm{M}_{1} \mathrm{M}_{4} \\ \mathrm{M}_{2} \mathrm{M}_{3} \\ \mathrm{M}_{2} \mathrm{M}_{4} \\ \mathrm{M}_{3} \mathrm{M}_{4} \end{array} \nonumber \]	\[\begin{array}{l} \mathrm{W}_{1} \mathrm{W}_{2} \mathrm{W}_{3} \\ \mathrm{W}_{1} \mathrm{W}_{2} \mathrm{W}_{4} \\ \mathrm{W}_{1} \mathrm{W}_{3} \mathrm{W}_{4} \\ \mathrm{W}_{2} \mathrm{W}_{3} \mathrm{W}_{4} \end{array} \nonumber \]

For every 2-person committee, four 3-person committees can be chosen to form a 5-person committee. If we choose \(M_1M_2\) as our 2-man committee, then we can choose any of \(W_1W_2W_3\), \(W_1W_2W_4\), \(W_1W_3W_4\), or \(W_2W_3W_4\) as our 3-woman committees. As a result, we get

\[ \boxed{M_1M_2}W_1W_2W_3, \boxed{M_1M_2} W_1W_2W_4, \boxed{M_1M_2} W_1W_3W_4, \boxed{M_1M_2} W_2W_3W_4 \nonumber \]

Similarly, if we choose \(M_1M_3\) as our 2-man committee, then, again, we can choose any of \(W_1W_2W_3\), \(W_1W_2W_4\), \(W_1W_3W_4\), or \(W_2W_3W_4\) as our 3-woman committees.

\[\boxed{M_1M_3}W_1W_2W_3, \boxed{M_1M_3} W_1W_2W_4, \boxed{M_1M_3}W_1W_3W_4, \boxed{M_1M_3}W_2W_3W_4 \nonumber \]

And so on.

Since there are six two-person committees, and for every two-person committee, there are four three-person committees, there are altogether \(6 \cdot 4 = 24\) five-person committees.

In essence, we are applying the multiplication axiom to the different combinations.

Example \(\PageIndex{2}\)

A high school club consists of 4 first-year students, five sophomores, five juniors, and six seniors. How many ways can a committee of 4 people be chosen that includes

1. One student from each class?
2. All juniors?
3. Two first-year students and two seniors?
4. No freshmen?
5. At least three seniors?

Solution

a. Applying the multiplication axiom to the combinations involved, we get

( ₄C₁ ) ( ₅C₁ ) ( ₅C₁ ) ( ₆C₁ ) = 600

b. We are choosing all four members from the five juniors, and none from the others.

₅C₄ = 5

c. ₄C₂ \(\cdot\) 6C2 = 90

d. Since we don't want any freshmen on the committee, we need to choose all members from the remaining 16. That is

₁₆C₄ = 1820

e. Of the four people on the committee, we want at least three seniors. This can be done in two ways. We could have three seniors and one non-senior, or all four seniors.

( ₆C₃ ) ( ₁₄C₁ ) + ₆C₄ = 295

Example \(\PageIndex{3}\)

How many five-letter word sequences consisting of 2 vowels and three consonants can be formed from the letters of the word INTRODUCE?

Solution

First, we select a group of five letters consisting of 2 vowels and three consonants.
Since there are four vowels and five consonants, we have

( ₄C₂ ) ( ₅C₃ )

Since our next task is to create word sequences from these letters, we multiply them by 5!.

( ₄C₂ ) ( ₅C₃ ) ( 5! ) = 7200.

Example \(\PageIndex{4}\)

A standard deck of playing cards consists of 52 cards, divided into four suits, each containing 13 cards. In how many different ways can a 5-card hand consisting of four cards of one suit and one of another be drawn?

Solution

We will solve the problem using the following steps.
Step 1. Select a suit.
Step 2. Select four cards from this suit.
Step 3. Select another suit.
Step 4. Select a card from that suit.

Applying the multiplication axiom, we have

Ways of selecting the first suit	Ways of selecting four cards from this suit	Ways of selecting the next suit	Ways of selecting a card from that suit
4C1	13C4	3C1	13C1

( ₄C1 ) ( ₁₃C₄ ) ( ₃C₁ ) ( ₁₃C₁ ) = 111,540.

A STANDARD DECK OF 52 PLAYING CARDS

As in the previous example, many examples and homework problems in this book refer to a standard deck of 52 playing cards. Before we end this section, we take a minute to describe a standard deck of playing cards, as some readers may not be familiar with this.

A standard deck of 52 playing cards consists of 4 suits, each with 13 cards.

Each suit is associated with a color, either black (spades, clubs) or red (diamonds, hearts)

Each suit contains 13 denominations (or values) for cards:

Nine numbers 2, 3, 4,…, 10 and Jack(J), Queen (Q), King (K), Ace (A).

The Jack, Queen, and King are called "face cards" because they have pictures on them. Therefore, a standard deck has 12 face cards: (3 values JQK ) x (4 suits ♦♥♠ ♣ )

We can visualize the 52 cards by the following display

Suit	Color	Values (Denominations)
♦ Diamonds	Red	2 3 4 5 6 7 8 9 10 J Q K A
♥ Hearts	Red	2 3 4 5 6 7 8 9 10 J Q K A
♠ Spades	Black	2 3 4 5 6 7 8 9 10 J Q K A
♣ Clubs	Black	2 3 4 5 6 7 8 9 10 J Q K A