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Combinations Calculator (nCr)
Combinations nCr Calculator The Combinations Calculator will find
the number of possible combinations
C(n, r) = ( ) =
n n! that can be obtained by taking a
r (r!(n − r)!) sample of items from a larger set.
Basically, it shows how many different
n choose r
possible subsets can be made from the
n (objects) = larger set. For this calculator, the order
of the items chosen in the subset does
r (sample) =
not matter.
Clear Calculate Factorial
There are n! ways of arranging
Answer:
n distinct objects into an
ordered sequence,
Get a Widget for this Calculator permutations where n = r.
© CalculatorSoup 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
where order does matter and repl
n
the set or population
r
subset of n or sample set
Combinations Formula:
n!
C(n, r) =
(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.
n!
C(n, r) =
(r!(n − r)!)
n!
C(n, 2) =
(2!(n − 2)!)
expanding the factorials,
1 × 2 × 3... × (n − 2) × (n − 1) × (n)
=
(2 × 1 × (1 × 2 × 3... × (n − 2)))
cancelling and simplifying,
(n − 1) × (n) n(n − 1)
= =
2 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.
Cite this content, page or calculator as:
Furey, Edward "Combinations Calculator (nCr)" at
https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php from
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