Kelly and Kyle are playing a card game, and Kyle is wondering why there never seem to be repeated hands. He figures that since there are only 52 cards in the deck, and each hand has five cards, there really should be more duplicate hands. After all, 5 is nearly 1/10 of 52.

Kelly tells him that she thinks there are thousands of possible combinations, and that she would be really surprised to see the same 5 card hand twice or more in a given game.

Who is correct? Why?

### Factorials and Combinations

Recall that a factorial of a positive integer *n* is the product of *n*, and all of the positive integers less than *n*. We write this as *n*! = *n*(*n* - 1)(*n* - 2) .. (3) (2) (1).

In order to develop the binomial theorem, we need to look at a related idea: combinations. If you have studied probability, you may be familiar with combinations and permutations. A **combination** is the number of ways you can choose *r* objects from a group of *n* objects, if the order of choosing does not matter. The following examples will help clarify the idea of a combination.

### Examples

#### Example 1

Earlier, you were asked a question about the disagreement between Kelly and Kyle.

Kelly thinks there are thousands of possible 5-card combinations in a deck of cards, Kyle thinks there should not be all that many.

This is a classic combinations problem in the form "52, choose 5":

Using the formula: \begin{align*}_{n}C_{r} = \binom{n} {r} = \frac{n!} {r!(n - r)!}\end{align*}

We get: \begin{align*}_{52}C_{5} = \binom{52} {5} = \frac{52!} {5!(52 - 5)!}\end{align*}

Simplify to: \begin{align*}\frac{52!} {5!(47)!}\end{align*}

Simplify more: \begin{align*}\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\end{align*}

And once more: \begin{align*}\frac{311,875,200}{120}\end{align*}

Gives: \begin{align*}2,598,960\end{align*}

Looks like Kelly undershot by quite a bit too!

#### Example 2

In a class of 20 students, 3 students are going to be chosen to form a committee to plan a fieldtrip. How many possible committees are there?

To answer this question, we need to figure out how many ways we can choose groups of 3 students from the 20 on the class. The order of choosing does not matter. That is, if I choose Amy, Juan, and Nina, it is the same as choosing Juan, then Amy, then Nina, or any other ordering of the three students.

In general, we can find the number of combinations of *r* objects chosen from *n* objects by the following:

\begin{align*}_{n}C_{r} = \binom{n} {r} = \frac{n!} {r!(n - r)!}\end{align*} |
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(Note that there are two different symbols for combinations: \begin{align*}_{n}C_{r}\end{align*} and \begin{align*}\binom{n} {r}\end{align*} You can use either one, though \begin{align*}_{n}C_{r}\end{align*} is what is used on the TI-83/84.

Therefore the number of combinations of 3 people from 20 people in the class is

\begin{align*}\frac{20!} {3!17!} = 1140\end{align*} |
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#### Example 3

How many different groups of 3 cards can be chosen from 20 different cards, assuming order does not matter? Use a graphing calculator.

To find \begin{align*}\binom{20} {3} \end{align*}:

Press: 20 <TI font_MATH> and then move right to the **PRB** menu.

Press 3. This takes you back to the main screen. You should see 20 \begin{align*}_{n}C_{r}\end{align*}.

Now press 3 >,TI font_ENTER>.

You should see the answer, 1140.

#### Example 4

Calculate by hand: How many different 4-person teams can be made from 7 people?

Smaller numbers, such as these, are not too difficult to calculate by hand.

\begin{align*}\binom{7} {4} = \frac{7!} {4!3!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} {4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6 \cdot 5} {3 \cdot 2 \cdot 1} = 7 \cdot 5 = 35\end{align*}.

Canceling factors in the numerator and denominator simplifies the calculation.

#### Example 5

Simplify: \begin{align*}\frac{(m + 3)!}{m + 1}!\end{align*}

First we need to expand the numerator and denominator:

\begin{align*}\frac{(m + 3)(m + 2)(m + 1)(m)(m - 1)...(1)}{(m + 1)(m)(m - 1)...(1)}\end{align*}

\begin{align*}(m + 3)(m + 2)\end{align*} ..... Cancel common factors

\begin{align*} m^2 + 5m +6\end{align*} ..... Simplify

#### Example 6

Demonstrate that: \begin{align*}_{2(4)}C_2 = 2(_4C_2) + 4^2\end{align*}.

Evaluate both sides:

\begin{align*}_8C_2 = 28\end{align*}

\begin{align*}_4C_2 = 6\end{align*}

\begin{align*} 4^2 = 16\end{align*}

Check: \begin{align*}28 = 2(6) + 16\end{align*}

\begin{align*}28 = 12 + 16\end{align*}

\begin{align*}28 = 28\end{align*} - the equality holds.

#### Example 7

In how many ways can I pick 6 jelly beans from a container containing 10 jelly beans?

We calculate using the formula: \begin{align*}_{10}C_6 = \frac{10!}{6!(10 - 6)!}\end{align*}

\begin{align*}10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)\end{align*}

\begin{align*}10! = 3628800\end{align*}

\begin{align*}6! = (1)(2)(3)(4)(5)(6)\end{align*}

\begin{align*}6! = 720\end{align*}

\begin{align*}(10 - 6)! = (1)(2)(3)(4)\end{align*}

\begin{align*}(10-6)! = 24\end{align*}

\begin{align*}\therefore _{10}C_6 = \frac{3628800}{720 \cdot 24} \to 210\end{align*}

#### Example 8

At the carnival, you decide to play a game of chance. You buy 15 tickets for the game. You have a 75% chance of winning each time you play the game. What is the probability that you will win exactly 8 of the 15 games?

The chance is about 4%.

For each of the 15 games, there is 75% chance of winning, and a 25% chance of losing.

The probability of exactly 8 wins is \begin{align*}\mathit \ \binom{15}{8}(.75)^{8}(.25)^{7}\approx 0.039 \to 3.9\%\end{align*}

Note that this is only the probability of winning exactly 8 games, no more, no less.

#### Example 9

Explain why the following equality holds: \begin{align*}_{(n+1)}C_r = _nC_r + _n C _{r-1}\end{align*}.

Proof of equality:

First, remove one item from the set.

Then, either the (r) item we want to come out of the remaining (n) items, or we choose (r - 1) item from the (n) remaining items and we include the one item we removed.

### Review

Simplify and evaluate the factorials.

- \begin{align*}\frac{(b - 2)!}{(b - 5)!}\end{align*}
- \begin{align*}\frac{(a + 2)!}{(a + 1)!}\end{align*}
- \begin{align*}\frac{7!}{3! 3!}\end{align*}
- \begin{align*}\frac{9!}{8!}\end{align*}
- \begin{align*} 4! + 3!\end{align*}
- \begin{align*}\frac {6!}{5!}\end{align*}

Show that the equality holds by evaluating both sides of the equation:

- \begin{align*}_4C_2 = _4C_{4-2}\end{align*}
- \begin{align*}_{(7+1)}C_4=_7C_4 + _7C_{(4-1)}\end{align*}

Explain conceptually why the following equality holds:

- \begin{align*}_{2n}C_2 = 2(_nC_2) + n^2\end{align*}

Solve.

- \begin{align*} _{8}C_{5}\end{align*}
- \begin{align*} _{6}C_{3}\end{align*}
- In a class of 200 students, 25 will be chosen randomly to participate in a research study. How many possible groups of 25 students can be chosen? (Hint: use a calculator!)
- A die is rolled 10 times. What is the probability of rolling exactly four 4’s? (Hint: the probability of rolling a 4 is 1/6. The probability of not rolling a 4 is 5/6.)
- The local TV station forecasts a 30% chance of rain every day for the next week. What is the probability that it will rain on exactly 6 out of the next 7 days?
- Consider the following situation: a basketball player is going to attempt to make 20 free throws. She is assuming that she has an 80% chance of making each shot. What is the probability that she will make exactly 19 out of 20 shots?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.11.