# Combinations without repetition

Find out how many different ways you can choose*k*items from

*n*items set. With/without repetition, with/without order.

## Calculation:

${C}_{k}(n)=\left({\displaystyle \genfrac{}{}{0px}{}{n}{k}}\right)={\displaystyle \frac{n!}{k!(n-k)!}}\text{}\phantom{\rule{0ex}{0ex}}\text{}\phantom{\rule{0ex}{0ex}}n=65\text{}\phantom{\rule{0ex}{0ex}}k=8\text{}\phantom{\rule{0ex}{0ex}}\text{}\phantom{\rule{0ex}{0ex}}{C}_{8}(65)=\left({\displaystyle \genfrac{}{}{0px}{}{65}{8}}\right)={\displaystyle \frac{65!}{8!(65-8)!}}={\displaystyle \frac{65\cdot 64\cdot 63\cdot 62\cdot 61\cdot 60\cdot 59\cdot 58}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}=5047381560$

### Number of combinations: 5047381560

5047381560

# A bit of theory - foundation of combinatorics

## Variations

A variation of the k-th class of n elements is an ordered k-element group formed from a set of n elements. The elements are not repeated and depend on the order of the group's elements (therefore arranged).The number of variations can be easily calculated using the combinatorial rule of product. For example, if we have the set n = 5 numbers 1,2,3,4,5 and we have to make third-class variations, their V

_{3}(5) = 5 * 4 * 3 = 60.

${V}_{k}(n)=n(n-1)(n-2)\mathrm{.}\mathrm{.}\mathrm{.}(n-k+1)={\displaystyle \frac{n!}{(n-k)!}}$

n! we call the factorial of the number n, which is the product of the first n natural numbers. The notation with the factorial is only clearer, equivalent. For calculations, it is fully sufficient to use the procedure resulting from the combinatorial rule of product.
## Permutations

The permutation is a synonymous name for a variation of the nth class of n-elements. It is thus any n-element ordered group formed of n-elements. The elements are not repeated and depend on the order of the elements in the group.$P(n)=n(n-1)(n-2)\mathrm{.}\mathrm{.}\mathrm{.}1=n!$

A typical example is: We have 4 books, and in how many ways can we arrange them side by side on a shelf?
## Variations with repetition

A variation of the k-th class of n elements is an ordered k-element group formed of a set of n elements, wherein the elements can be repeated and depends on their order. A typical example is the formation of numbers from the numbers 2,3,4,5, and finding their number. We calculate their number according to the combinatorial rule of the product:${V}_{k}^{\mathrm{\prime}}(n)=n\cdot n\cdot n\cdot n\mathrm{.}\mathrm{.}\mathrm{.}n={n}^{k}$

## Permutations with repeat

A repeating permutation is an arranged k-element group of n-elements, with some elements repeating in a group. Repeating some (or all in a group) reduces the number of such repeating permutations.${P}_{{k}_{1}{k}_{2}{k}_{3}\mathrm{.}\mathrm{.}\mathrm{.}{k}_{m}}^{\mathrm{\prime}}(n)={\displaystyle \frac{n!}{{k}_{1}!{k}_{2}!{k}_{3}!\mathrm{.}\mathrm{.}\mathrm{.}{k}_{m}!}}$

A typical example is to find out how many seven-digit numbers formed from the numbers 2,2,2, 6,6,6,6.
## Combinations

A combination of a k-th class of n elements is an unordered k-element group formed from a set of n elements. The elements are not repeated, and it does not matter the order of the group's elements. In mathematics, disordered groups are called sets and subsets. Their number is a combination number and is calculated as follows:${C}_{k}(n)=\left({\displaystyle \genfrac{}{}{0px}{}{n}{k}}\right)={\displaystyle \frac{n!}{k!(n-k)!}}$

A typical example of combinations is that we have 15 students and we have to choose three. How many will there be?
## Combinations with repeat

Here we select k element groups from n elements, regardless of the order, and the elements can be repeated. k is logically greater than n (otherwise, we would get ordinary combinations). Their count is:${C}_{k}^{\mathrm{\prime}}(n)=\left({\displaystyle \genfrac{}{}{0px}{}{n+k-1}{k}}\right)={\displaystyle \frac{(n+k-1)!}{k!(n-1)!}}$

Explanation of the formula - the number of combinations with repetition is equal to the number of locations of n − 1 separators on n-1 + k places.
A typical example is: we go to the store to buy 6 chocolates. They offer only 3 species. How many options do we have?
k = 6, n = 3.
## Foundation of combinatorics in word problems

- PIN - codes

How many five-digit PIN - code can we create using the even numbers? - Mastering

The student masters the subject matter for the exam in Czech to 98%, from Math to 86% and from Economics to 71%. What is the probability that he will fail from Math and from others will succeed? - Six questions test

There are six questions in the test. There are 3 answers to each - only one is correct. In order for a student to take the exam, at least four questions must be answered correctly. Alan didn't learn at all, so he circled the answers only by guessing. What - MATES

In MATES (Small Television tipping), from 35 random numbers drawn 5 winning numbers. How many possible combinations are there? - Three dice

The player throwing the three dice asked G. Galilei the question: "Should I bet on the sum of 11 or the sum of 12?" What did Galilei answer him? Hint: write down all three triples of numbers that can be thrown and: have a total of 11 have a total of 12 an - 7 heroes

9 heroes galloping on 9 horses behind. How many ways can sort them behind? - Blocks

There are nine interactive basic building blocks of an organization. How many two-blocks combinations are there? - Distribution function

X 2 3 4 P 0.35 0.35 0.3 The data in this table do I calculate the distribution function F(x) and then probability p(2.5 < ξ < 3.25) p(2.8 < ξ) and p(3.25 > ξ) - Chords

How many 4-tones chords (chord = at the same time sounding different tones) is possible to play within 7 tones? - Tokens

In the non-transparent bags are red, white, yellow, blue tokens. We 3times pull one tokens and again returned it, write down all possibilities. - In the orchard

In the orchard, they planted 25 apple trees, 20 pears, 15 plums and 40 marbles. A strong late frost, however, destroyed a fifth of all new trees. Unfortunately, it was all the trees of one kind of fruit. What is the probability that the plums have died ou - Three subjects

In a class of 40 students, 18 passed mathematics, 19 passed accounts, 16 passed economics, 5 mathematics and accounts only, 6 mathematics only, 9 accounts only, 2 accounts and economics only, if each student offered at least one of the subjects. a) how ma - Ace

From complete sets of playing cards (32 cards), we pulled out one card. What is the probability of pulling the ace? - Hockey match

The hockey match ended with a result of 3:1. How many different storylines may have the match? - Kids

How many different ways can sit 8 boys and 3 girls in line if girls want to sit on the edge? - Sweater

Dana confuses sweater and wool has a choice of seven colors. In how many ways can she choose from three colors to the sleeves?

next math problems »