## Calculator Use

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:

*C ^{R}(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.

## FAQs

### How do you calculate nCr combination? ›

We utilise the nCr formula to compute combinations. This formula is as follows: **nCr = n! / r!** *** (n – r)!**, where n is the total number of items and r is the number of things that may be selected at one time.

**How do you calculate how many possible combinations? ›**

The number of combinations of n objects taken r at a time is determined by the following formula: **C(n,r)=n!** **(n−r)!**

**How many combinations can you make with 4 numbers? ›**

There are **10,000** possible combinations that the digits 0-9 can be arranged into to form a four-digit code.

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

(n - r)! 5C_{3} = 5!/ [3! (5 - 3)!] Therefore, there are **10** ways to choose a committee of 3 people from a group of 5 people.

**How many possible combinations of 4 numbers without repeating? ›**

The number of possible combinations with 4 numbers without repetition is 15. The formula we use to calculate the number of n element combinations when repetition is not allowed is **2n - 1**.

**How many combinations does a 33x33 have? ›**

The original 3x3x3 Rubik's cube has **43 252 003 274 489 856 000** combinations, or 43 quintillion. Again, as pointed out on this website's main page, this is a manageably imaginable number.

**How many combinations are there from 1 to 30? ›**

Answer and Explanation: The number of combinations that are possible with 30 numbers, where numbers can't be repeated, is **1,073,741,823**.

**How do you find the number of combinations without repetition? ›**

Combinations are selections of objects, with or without repetition, order does not matter. The number of k-element combinations of n objects, without repetition is Cn,**k = (n k ) = n! k!(** **n − k)!** .

**How many numbers can you make out of 1234? ›**

No. of numbers that can be formed using all four digits 1,2,3,4=**4!**

**How many times can 1234 be arranged? ›**

You multiply all the elements together to get the total number of possible combinations. Thus, 1*2*3*4 =24, which is indeed the total # of possible combinations for this number set.

### How many 4 digit numbers can you make 1234? ›

using the digits 1234 without repetition four digit numbers can be formed there are **24** of such 4 digit numbers what is the average of this 24 number See what the community says and unlock a badge.

**How many combinations of 4 people can be formed from 6 people? ›**

So, the number of combinations of size 4 can be formed from a set of 6 distinct objects is equal to **15** .

**How many combinations of 3 people can be formed from 6 people? ›**

So, 3 team members from 6 students can be formed in **20 ways**.

**How many ways can we choose 2 students from a group of 5? ›**

In both of our solving processes, we see that 5 C 2 = 10. In other words, there are **10 possible combinations of 2 objects chosen from 5 objects.**

**How many combinations of 7 numbers are there? ›**

Answer and Explanation: The number of combinations that are possible with 7 numbers is **127**. In general, the formula we use to determine the number of combinations possible with n elements is as follows: Number of combinations possible with n elements = 2n - 1.

**What is the formula for combination? ›**

Formula for Combination

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

**How many 5 digit numbers can be formed from 12345 without repetition? ›**

Not a single five-digit prime number can be formed using the digits **1, 2, 3, 4, 5**(without repetition). This is because if one adds the digits, the result obtained will be = 1 + 2 + 3 + 4 + 5 = 15 which is divisible by 3.

**What is the probability of guessing a 4 digit PIN code? ›**

Thus, if you picked a 4-digit number randomly, you'd have a **one in 10,000** chance of picking that number (or any other specific 4-digit number).

**How many 4 digit numbers can be formed from 1234 and 5 if no repetition is allowed? ›**

Summary: The number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated is **120**.

**What number is 43252003274489856000? ›**

43,252,003,274,489,856,000 **represents the number of possible configurations for a single Rubik's Cube**. Mind-blowing, isn't it? Like it or not, the Rubik's Cube has mesmerized millions of people around the world, and has spawned a whole breed of niche clubs, obscure competitions, and even me-too products.

### How many scrambles on a 19x19? ›

On 19x19, there are about **10 ^{170}** possible positions, but a rough estimate of 361!

**How many scrambles on a 7x7? ›**

The 7x7x7 Rubik's cube (the V-Cube 7) has 19 500 551 183 731 307 835 329 126 754 019 748 794 904 992 692 043 434 567 152 132 912 323 232 706 135 469 180 065 278 712 755 853 360 682 328 551 719 137 311 299 993 600 000 000 000 000 000 000 000 000 000 000 000 combinations.

**How many combinations of 7 numbers from 1 to 50? ›**

In other words, there are **5040** different ways that the 7 numbers you choose can be filled out on your lottery ticket--if you choose your 7 numbers correctly, any of these ways will make a winning ticket.

**What are the 24 combinations of 1234? ›**

**1234, 1243, 1423, 4123, 1324, 1342, 1432, 4132, 3124, 3142, 3412, 4312, 2134, 2143, 2413, 4213, 2314, 2341, 2431, 4231, 3214, 3241, 3421, 4321**.

**How many combinations of 20 items are there? ›**

Answer and Explanation: The number of combinations that are possible with 20 numbers is **1,048,575**.

**How many 5 digit combinations are there no repeats? ›**

Thus you have made 5 × 4 × 3 × 2 1 = **120** choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated.

**How many combinations of 5 items are there without repeating? ›**

Note that your choice of 5 objects can take any order whatsoever, because your choice each time can be any of the remaining objects. So we say that there are 5 factorial = 5! = 5x4x3x2x1 = **120** ways to arrange five objects.

**How many combinations are there between 0000 and 9999? ›**

So **5040** different numbers can be made from 0000 to 9999 without repetitions.

**How many numbers can 1234567890 make? ›**

Hello there, The answer is known as a factorial, which for the number 1234567890 of 10 total digists = 10! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3,628,800 - **a little under 3.3 million** possible combinations.

**How many 4 digit combinations are there with 1,2 numbers? ›**

In your case, with 12 numbers, the number is 12x11x10x... x2x1=479001600. This number is called "twelve factorial" and written 12!, so, for example 4!= 4x3x2x1=**24**.

### How many 4 digit combinations are there using 1 6? ›

6 x 5 x 4 x 3 = **360** possible combinations. Given a 4 digit integer. each digit can be 1 through 6.

**How 6174 is magic number? ›**

This number is renowned for the following rule: Take any four-digit number, using at least two different digits (leading zeros are allowed). Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary. Subtract the smaller number from the bigger number.

**How many combinations of 24 are there? ›**

Answer and Explanation: The number of combinations possible with 24 numbers is **16,777,215**.

**How easy is it to crack a 4 digit code? ›**

**You can crack more than 10 percent of random PINs by dialing in 1234**. Expanding a bit, 1234, 0000, and 1111, make up about 20 percent. 26.83 percent of passwords can be cracked using the top 20 combinations. That would be 0.2 percent of the passwords if they were randomly distributed.

**How long would it take to guess a 4 digit code? ›**

**If a password is only four or five characters (whether they are just numbers or a combination of numbers, letters and symbols), there's a very high chance that it will be hacked instantly**. However, if a password is only numbers and up to 18 characters, it could take a hacker up to nine months to crack the code.

**What are the most common 4 digit number combinations? ›**

The most common four-digit PINs, according to the study, are **1234, 0000, 2580 (the digits appear vertically below each other on the numeric keypad), 1111 and 5555**.

**How many combinations of 4 people can be formed from 7 people? ›**

Hence, with the help of the formula (A) and (B) we have determined the required number of ways a committee of 4 people can be selected from a group of 7 people is **35**.

**How many ways can 4 people be chosen from a group of 9? ›**

In **126** ways a committee of 4 be chosen from a group of 9 people.

**How many groups of 3 can be made from 8? ›**

n=8, r=3 n = 8 , r = 3 . Therefore, **56 committees** of 3 people can be formed from 8 people.

**How many ways can you pick 3 out of 10? ›**

P(10,3) = **720**.

### How many combinations of 3 out of 5? ›

So 5 choose 3 = **10** possible combinations.

**How many ways a 3 seater can be filled with 5 persons? ›**

=5P3=(5−3)! 5! =2×15×4×3×2×1=**60**.

**How many ways 4 students can be chosen from the class of 20 students *? ›**

=24, meaning there are **23 unneeded permutations** for each permutation we computed.

**How do you calculate how many combinations are possible? ›**

The number of combinations of n objects taken r at a time is determined by the following formula: **C(n,r)=n!** **(n−r)!**

**How many ways can you select a committee of 5 students out of 10 students? ›**

Therefore, the number of ways of selecting a committee of 5 members from a group of 10 persons is **252**.

**How do you calculate nCr and nPr? ›**

**nCr = n!/(r!*(** **n-r)!)** **nPr = n!/(n-r)!**.

**What does nCr calculate? ›**

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. There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.

**What is the rule of nCr? ›**

**C r n = n !** **r !** **n - r !** , where n is the total number of elements and r is the number of elements that have to be selected.

**How do you calculate nCr fast? ›**

To calculate combinations we use the nCr formula: **nCr = n! / r!** *** (n - r)!**, where n = number of items, and r = number of items being chosen at a time.

**What is nCr in Desmos? ›**

nCr(n, r) **Number of Combinations**, where n is the total number of items, and r is the number in the combination. nPr(n, r) Number of Permutations, where n is the total number of items, and r is the number in the permutation.

### How to solve 7c3? ›

Here ${}^7{C_3} $ means on **selecting the 3 things out of 7 things gives the value 35**.