# Combinations

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##### Examples

###### Lessons

- Compare the difference between the following 2 scenarios:
- How many ways can a president, a vice president, and a treasurer be selected from a class of 20 students?
- How many different committees of 3 people can be selected from a class of 20 students?

- A standard deck of 52 cards consists of:

• 4 suits: diamonds, hearts, spades, and clubs

• each suit has 13 cards

• red cards: diamonds and hearts

• black cards: spades and clubs

• face cards: Jacks, Queens, and Kings

How many different 5-card hands can be formed containing: **Problems with "at least", "at most"**

From a standard deck of 52 cards, how many different 5-card hands can be formed containing:- There are 20 members in a student council, 7 boys and 13 girls. How many ways can a committee of 5 people be selected if there must be:
- Five points are marked on a circle. By connecting the points on the circle,

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###### Topic Notes

How many choices do I have? This lesson can help you with answering this question! Combination is the process of selecting members from a set of items. Combination formula is your best friend in this lesson. Unlike permutations, order doesn't matter in combinations!

## Combinations

As you know, we have already talked about the concepts and comparison of permutations vs combinations in our last lesson, still, since these are the two most important processes used in combinatorics, we are dedicating a specific lesson for each of them. Today is the case for combinations!

Let us start with a small review:

#### How to calculate combinations

We have already learnt that combinations are a way to select objects coming from a complete set, and while permutations are arrangements (ordering) of items, combinations just focus on the specific items selected but does not care about how they are arranged.

To demonstrate how combinations work, we use a set of four boxes:

You are asked to find how many possible combinations can you do with two (any two) boxes out of these four.

Notice that in permutations, the pairs of A B and B A are counted as different even when you have the exact same two letters involved; the same goes for the case of B C and C B, D B and B D, A C and C A, etc. This is the main difference we were referencing before when comparing combinations vs permutations, since in combinations, A B and B A are counted as one combination only! Why? Because a combination is taken as a selection of two distinct items from the set, no matter the order in which they are arranged; thus affirming, that combinations are about how we select items from a set, not how we arrange them.

A formal combination definition would be that when selecting items from a set, each combination is a group (the size of the group depending on what is asked) consisting of distinct items from a set no matter in which order they are arranged. Therefore, when using two out of the four boxes shown in figure 4 we have the following:

Where you can clearly see the difference between a permutation and a combination, and notice how the number of permutations for the same set and conditions is higher than that of combinations; this is because permutations double count combinations.

With that in mind, the number of combinations formula is defined as:

$\quad$Where:

$\quad \large _{n}C_{r}$ = Combinations of $r$ items from a set of $n$ items

$\qquad n$ = number of total distinct items in a set

$\qquad r$ = items selected from the set

Notice the combinations formula shown in equation 1 is very similar to the permutations formula shown in earlier lessons, the difference comes from an extra factorial in the denominator, which takes care of eliminating the double counting of the permutations.

And so, before we move on to problem examples, let us see a few conclusions: Permutations are about arranging or ordering items obtained from a set, combinations are about selecting distinct items from a set. When looking at the results from combinations and permutations, the main difference between a permutation and combination is that the first pays attention to the order in which the subset of items picked from a total set is arranged, while how many combinations can be done in a subset are characterized by not noticing the way the objects are arranged, just the distinction between the ones selected (no matter how they are placed into positions).

#### Combinatorics problems

We have named this section as combinatorics problems because we will start with a problem in which permutations and combinations are compared, still, the rest of the problems are focused on our topic for today: combinations.

__Example 1__

Compare the difference between the following 2 scenarios:
**1. $\quad$ How many ways can a president, a vice president, and a treasurer be selected from a class of 20 students?**

Therefore, there are 6,840 different ways in which the class cabinet can be selected in a class of 20 students, where we have distinction between who ends up in each position from the committee. Now take a look at the next question to see the difference:

**2. $\quad$ How many different committees of 3 people can be selected from a class of 20 students?**

As you can see, this problem has been designed so you can clearly see the difference between permutations and combinations, and even, between the permutations and combinations formula! Once more, the difference relies on the importance of position arrangement for permutations, while the formula for combinations eliminates this by adding a factorial multiple on its denominator.

__Example 2__

A standard deck of 52 cards consists of:
- 4 suits: diamonds, hearts, spades, and clubs
- each suit has 13 cards
- red cards: diamonds and hearts
- black cards: spades and clubs
- face cards: Jacks, Queens, and Kings

**1. $\quad$ Any 5 cards (no restrictions)?**

Without any restrictions, we have more than two and a half million combinations!

Before we pass to the next part of this problem, notice that the restrictions we are talking about, do not refer to the order of the cards but they could talk about the colors, suits, numbers or ranks, all of which can be taken into account in case of selecting a hand. For this reason, all of the questions on this example problem are dealt with the combination formula, since the order in which the cards are arranged in a hand has no importance, the distinctions on the cards are made based on their characteristics and once they have been picked in a hand, we dont care in which order they are arranged.

**2. $\quad$ All black cards?**

**3. $\quad$ 1 black card and 4 red cards?**

And for choosing 4 red cards we have that $n$ = 26 and $r$ = 4, therefore:

Therefore, the total amount of combinations for 1 black and 4 red cards in a 5-card hand, is equal to the product of the combinations found above:

*possible combinations of 1 black and 4 red*= 26 × 14,950 = 388,700

**4. $\quad$ All face cards?**

**5. $\quad$ 3 Jacks and 2 Kings?**

Let us start with the combinations for the Jacks, we need 3 of them out of 4, and so we have that $n$ = 4 and $r$ = 3, therefore:

And for the two Kings we have that $n$ = 4 and $r$ = 2, therefore:

To obtain the total amount of possible combinations of 3 Jacks and 2 Kings in a 5-card hand, we multiply the combinations obtained above and have that:

*possible combinations of 3 Jacks and 2 Kings*= 4 × 6 = 24

**6. $\quad$ 3 Jacks?**

The possible combinations for 3 Jacks in a hand of cards were obtained above in equation 10, and so we know there are 4 possible combinations for this to happen.

Now, let us calculate combinations coming from the rest of the 48 possible cards to land in the 2 positions left in our 5-card hand, for which we have that $n$ = 48 and $r$ = 2, therefore:

And so, putting these combined results, we can obtain the total amount of possible combinations of 5-card hands when there are 3 jacks:

*possible combinations of 3 Jacks in a hand*= 4 × 1,128 = 4,512

__Example 3__

On this example will take a look at problems using phrases such as "at least" and "at most", such problems can be confusing as to what are the conditions and how to tackle them, so check it out:
From a standard deck of 52 cards, how many different 5-card hands can be formed containing:

**1. $\quad$ Exactly 2 Diamonds?**

Calculating the possible combinations of two diamond cards out of 13:

Now for the possible combinations of any of the 39 cards left, to fill the 3 positions left in the card hand:

And so, the total amount of combinations for exactly 2 diamonds in a 5-card hand is:

*combinations for 2 diamonds on a hand*$\, = \, _{13}C_{2} \times \, _{39}C_{3} =$ 78 × 9,139 = 712,842

The answer to part 1 of example problem 3 is that we have a total of 712,842 possible combinations for 5-card hands with exactly 2 diamond cards.

In the same way we calculated the combinations for exactly 2 diamonds, we can calculate the total amount of combinations for having exactly 0, 1, 3, 4, and up to 5 diamond cards in a 5-card hand:

*combinations for 0 diamonds on a hand*$\, = \, _{13}C_{0} \times \, _{39}C_{5} =$ 1 × 575,757 = 575,757

*combinations for 1 diamonds on a hand*$\, = \, _{13}C_{1} \times \, _{39}C_{4} =$ 13 × 82,251 = 1,069,263

*combinations for 2 diamonds on a hand*$\, = \, _{13}C_{2} \times \, _{39}C_{3} =$ 78 × 9,139 = 712,842

*combinations for 3 diamonds on a hand*$\, = \, _{13}C_{3} \times \, _{39}C_{2} =$ 286 × 741 = 211,926

*combinations for 4 diamonds on a hand*$\, = \, _{13}C_{4} \times \, _{39}C_{1} =$ 715 × 39 = 27,885

*combinations for 5 diamonds on a hand*$\, = \, _{13}C_{5} \times \, _{39}C_{0} =$ 1,287 × 1 = 1,287

Notice on equation 18 we mentioned the combinations for two diamond cards in a hand again. Why is this important to know? Because as we have said before, this problem is focused in questions containing phrases at least and at most, and so, we will need these quantities to respond the next two questions.

Keep in mind we already know that all the possible 5-card hands from a 52-card deck are equal to:

We calculated this in the first part of our last problem (equation 4).

Now you have all you need to respond to the next two questions!

**2. $\quad$ At least 2 Diamonds?**

*combinations for 2 diamonds on a hand*$\, = \, _{13}C_{2} \times \, _{39}C_{3} =$ 712,842

*combinations for 3 diamonds on a hand*$\, = \, _{13}C_{3} \times \, _{39}C_{2} =$ 211,926

*combinations for 4 diamonds on a hand*$\, = \, _{13}C_{4} \times \, _{39}C_{1} =$ 27,885

*combinations for 5 diamonds on a hand*$\, = \, _{13}C_{5} \times \, _{39}C_{0} =$ 1,287

And so, to find all of the possible combinations of cards in a hand with at least 2 diamonds, we need to add the four quantities shown in equation 20:

Our final answer is: 953,940 combinations with at least 2 diamond cards.

**3. $\quad$ At most 2 Diamonds?**

*combinations for 0 diamonds on a hand*$\, = \, _{13}C_{0} \times \, _{39}C_{5} =$ 575,757

*combinations for 1 diamonds on a hand*$\, = \, _{13}C_{1} \times \, _{39}C_{4} =$ 1,069,263

*combinations for 2 diamonds on a hand*$\, = \, _{13}C_{2} \times \, _{39}C_{3} =$ 712,842

Therefore, we add the results for each of them to find the total we are looking for:

Our final answer is: 2,357,862 combinations with 2 diamond cards at the most.

There are two more examples included on this lessons videos that you should check up, one of them focuses in problems for at least and at most scenarios.

This is the end of our lesson, we recommend you to check out this link on permutations and combinations, which will provide with a simple and summarized explanation of the two in preparation of our last lesson on combinatorics.

We hope you enjoyed this topic, and see you in the next one!

Combination: nCr = number of selections of r items taken from a set of n distinct items (order does NOT matter!!)

= $\frac{{n!}}{{(n - r)!\;\;r!}}$

= $\frac{{n!}}{{(n - r)!\;\;r!}}$

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