How many different combinations are possible when I choose a group of 3 people from 5 people?

Rhonda J.

asked • 01/02/14

Word problem that has me confused

4 Answers By Expert Tutors

Robert J. answered • 01/02/14

Certified High School AP Calculus and Physics Teacher

Since the order doesn't matter in this problem, there are 5C3 = 5C2 = 5*4/2 = 10 different groups of 3 that can be picked.

Grigori S. answered • 01/03/14

Certified Physics and Math Teacher G.S.

This is the number of combinations of 3 elements out of 5.

Example.If you are given 5 digits 1,2,3,4,5 you can pick up 3 elements in the following ways:

123  124  125   234   235   345   134    135   245    145

Because the order in which they are picked up doesn't matter (123 is the same collection like 321 or 231)

these are all possible combinations, 10. Robert has shown you the direct way of calculations:

this is a combination problem

you are choosing 3 people from a group of 5 people

the formula is n!/[(r!)(n-r)!] where n is the number of people(5) and r is how many you are choosing(3)

! means factorial: for example 7!=7*6*5*4*3*2*1

so we have (5*4*3*2*1)/[(3*2*1)(2*1)] because n-r=5-3=2

notice that the numerator and the denominator both have 3*2*1 in them, so they cancel out and you have

if you did not cancel or simplify you would have 120/(6)(2)=120/12=10 again

Steve S. answered • 01/02/14

Tutoring in Precalculus, Trig, and Differential Calculus

5*4*3 = number of groups if order matters3*2*1 = number of ways one group can be ordered5*4*3 / (3*2*1) = number of groups if order does not matter= 5*2*2*3*1 / (3*2*1)= 5*2 = 10Where do the factorials come from in the formula for mCn?5*4*3 = 5*4*3*2*1 / (2*1) = 5! / 2!3*2*1 = 3!

5*4*3 / (3*2*1) = 5! / 2! / 3! = 5! / (2! * 3!) = 5C3 = 5C2

Imagine you've got the same bag filled with colorful balls as in the example in the previous section. Again, you pick five balls at random, but this time, the order is important - it does matter whether you pick the red ball as first or third. Let's take a more straightforward example where you choose three balls called R(red), B(blue), G(green). There are six permutations of this set (the order of letters determines the order of the selected balls): RBG, RGB, BRG, BGR, GRB, GBR, and the combination definition says that there is only one combination! This is the crucial difference.

By definition, a permutation is the act of rearrangement of all the members of a set into some sequence or order. However, in literature, we often generalize this concept, and we resign from the requirement of using all the elements in a given set. That's what makes permutation and combination so similar. This meaning of permutation determines the number of ways in which you can choose and arrange r elements out of a set containing n distinct objects. This is called r-permutations of n (sometimes called variations). The permutation formula is as below:

P(n,r) = n!/(n-r)!.

Doesn't this equation look familiar to the combination formula? In fact, if you know the number of combinations, you can easily calculate the number of permutations:

P(n,r) = C(n,r) * r!.

If you switch on the advanced mode of this combination calculator, you will be able to find the number of permutations.

You may wonder when you should use permutation instead of a combination. Well, it depends on whether you need to take order into account or not. For example, let's say that you have a deck of nine cards with digits from 1 to 9. You draw three random cards and line them up on the table, creating a three-digit number, e.g., 425 or 837. How many distinct numbers can you create?

P(9,3) = 9!/(9-3)! = 9!/6! = 504

Check the result with our nCr calculator! And how many different combinations are there?

C(9,3) = 9!/(3! * (9-3)!) = 9!/(3! * 6!) = 84

The number of combinations is always smaller than the number of permutations. This time, it is six times smaller (if you multiply 84 by 3! = 6, you'll get 504). It arises from the fact that every three cards you choose can be rearranged in six different ways, just like in the previous example with three color balls.

Both combination and permutation are essential in many fields of learning. You can find them in physics, statistics, finances, and of course, math. We also have other handy tools that could be used in these areas. Try this log calculator that quickly estimate logarithm with any base you want and the significant figures calculator that tells you what are significant figures and explains the rules of significant figures. It is fundamental knowledge for every person that has a scientific soul.

How many different combinations of groups of $3$ can I make of $6$ people: $A, B, C, D, E,$ and $F$?

{(group1), (group2), (group3)}

1. The order of each group doesn't matter
2. The order of the whole set of groups doesn't matter either

One combination:

$(A,B), (C,D), (E,F)$

Another valid combination

$(A,B), (C,E), (D,F)$

The order of the 3 groups doesn't matter so:

This combination: $(A,B), (C,E), (D,F)$, is the same as: $(C,E), (D,F), (A,B)$

The order of each of the 3 groups consisting members doesn't matter too.

This combination: $(A,B), (C,E), (D,F)$, is the same as: $(B, A), (E, C), (F,D)$

What formula can be used to determine this?

The Combinations Calculator will find the number of possible combinations that can be obtained by taking a sample of items from a larger set. Basically, it shows how many different possible subsets can be made from the larger set. For this calculator, the order of the items chosen in the subset does not matter.

Factorial There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r. Combination The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed. Permutation The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed.  When n = r this reduces to n!, a simple factorial of n. Combination Replacement The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed. Permutation Replacement The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed. n the set or population r subset of n or sample set

Combinations Formula:

\( C(n,r) = \dfrac{n!}{( r! (n - r)! )} \)

For n ≥ r ≥ 0.

The formula show us the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects where order does not matter and repetitions are not allowed. [1] "The number of ways of picking r unordered outcomes from n possibilities." [2]

Also referred to as r-combination or "n choose r" or the binomial coefficient.  In some resources the notation uses k instead of r so you may see these referred to as k-combination or "n choose k."

Combination Problem 1

Choose 2 Prizes from a Set of 6 Prizes

You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6. How many different combinations of 2 prizes could you possibly choose?

In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.”

C (6,2)= 6!/(2! * (6-2)!) = 6!/(2! * 4!) = 15 Possible Prize Combinations

The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6}

Combination Problem 2

Choose 3 Students from a Class of 25

A teacher is going to choose 3 students from her class to compete in the spelling bee. She wants to figure out how many unique teams of 3 can be created from her class of 25.

In this example, we are taking a subset of 3 students (r) from a larger set of 25 students (n). Looking at the formula, we must calculate “25 choose 3.”

C (25,3)= 25!/(3! * (25-3)!)= 2,300 Possible Teams

Combination Problem 3

Choose 4 Menu Items from a Menu of 18 Items

A restaurant asks some of its frequent customers to choose their favorite 4 items on the menu. If the menu has 18 items to choose from, how many different answers could the customers give?

Here we take a 4 item subset (r) from the larger 18 item menu (n). Therefore, we must simply find “18 choose 4.”

C (18,4)= 18!/(4! * (18-4)!)= 3,060 Possible Answers

Handshake Problem

In a group of n people, how many different handshakes are possible?

First, let's find the total handshakes that are possible. That is to say, if each person shook hands once with every other person in the group, what is the total number of handshakes that occur?

A way of considering this is that each person in the group will make a total of n-1 handshakes. Since there are n people, there would be n times (n-1) total handshakes. In other words, the total number of people multiplied by the number of handshakes that each can make will be the total handshakes. A group of 3 would make a total of 3(3-1) = 3 * 2 = 6. Each person registers 2 handshakes with the other 2 people in the group; 3 * 2.

Total Handshakes = n(n-1)

However, this includes each handshake twice (1 with 2, 2 with 1, 1 with 3, 3 with 1, 2 with 3 and 3 with 2) and since the orginal question wants to know how many different handshakes are possible we must divide by 2 to get the correct answer.

Total Different Handshakes = n(n-1)/2

Handshake Problem as a Combinations Problem

We can also solve this Handshake Problem as a combinations problem as C(n,2).

n (objects) = number of people in the group
r (sample) = 2, the number of people involved in each different handshake

The order of the items chosen in the subset does not matter so for a group of 3 it will count 1 with 2, 1 with 3, and 2 with 3 but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively.

\( C(n,r) = \dfrac{n!}{( r! (n - r)! )} \)

\( C(n,2) = \dfrac{n!}{( 2! (n - 2)! )} \)

expanding the factorials,

\( = \dfrac{1\times2\times3...\times(n-2)\times(n-1)\times(n)}{( 2\times1\times(1\times2\times3...\times(n-2)) )} \)

cancelling and simplifying,

\( = \dfrac{(n-1)\times(n)}{2} = \dfrac{n(n-1)}{2} \)

which is the same as the equation above.

Sandwich Combinations Problem

This is a classic math problem and asks something like How many sandwich combinations are possible? and this is how it generally goes.

Calculate the possible sandwich combinations if you can choose one item from each of the four categories:

• 1 bread from 8 options
• 1 meat from 5 options
• 1 cheese from 5 options
• 1 topping from 3 options

Often you will see the answer, without any reference to the combinations equation C(n,r), as the multiplication of the number possible options in each of the categories. In this case we calculate:

8 × 5 × 5 × 3 = 600
possible sandwich combinations

In terms of the combinations equation below, the number of possible options for each category is equal to the number of possible combinations for each category since we are only making 1 selection; for example C(8,1) = 8, C(5,1) = 5 and C(3,1) = 3 using the following equation:

C(n,r) = n! / ( r!(n - r)! )

We can use this combinations equation to calculate a more complex sandwich problem.

Sandwich Combinations Problem with Multiple Choices

Calculate the possible combinations if you can choose several items from each of the four categories:

• 1 bread from 8 options
• 3 meats from 5 options
• 2 cheeses from 5 options
• 0 to 3 toppings from 3 options

Applying the combinations equation, where order does not matter and replacements are not allowed, we calculate the number of possible combinations in each of the categories. You can use the calculator above to prove that each of these is true.

• 1 bread from 8 options is C(8,1) = 8
• 3 meats from 5 options C(5,3) = 10
• 2 cheeses from 5 options C(5,2) = 10
• 0 to 3 toppings from 3 options; we must calculate each possible number of choices from 0 to 3 and get C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8

Multiplying the possible combinations for each category we calculate:

8 × 10 × 10 × 8 = 6,400
possible sandwich combinations

How many possible combinations are there if your customers are allowed to choose options like the following that still stay within the limits of the total number of portions allowed:

• 2 portions of one meat and 1 portion of another?
• 3 portions of one meat only?
• 2 portions of one cheese only?

In the previous calculation, replacements were not allowed; customers had to choose 3 different meats and 2 different cheeses. Now replacements are allowed, customers can choose any item more than once when they select their portions. For meats and cheeses this is now a combinations replacement or multichoose problem using the combinations with replacements equation:

CR(n,r) = C(n+r-1, r) = (n+r-1)! / (r! (n - 1)!)

For meats, where the number of objects n = 5 and the number of choices r = 3, we can calculate either combinations replacement CR(5,3) = 35 or substitute terms and calculate combinations C(n+r-1, r) = C(5+3-1, 3) = C(7, 3) = 35.

Calculating cheese choices in the same way, we now have the total number of possible options for each category at

• bread is 8
• meat is 35
• cheese is 15
• toppings is 8

and finally we multiply to find the total

8 × 35 × 15 × 8 = 33,600
possible sandwich combinations!

How many combinations are possible if customers are also allowed replacements when choosing toppings?

References

[1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 206, 2003.

For more information on combinations and binomial coefficients please see Wolfram MathWorld: Combination.