# Combinations without repetition n=11, k=3 result

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

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

## Calculation:

$C_{k}(n)=(kn )=k!(n−k)!n! n=11k=3C_{3}(11)=(311 )=3!(11−3)!11! =3⋅2⋅111⋅10⋅9 =165$

### The number of combinations: 165

# A bit of theory - the 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)...(n−k+1)=(n−k)!n! $

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 and 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.$P(n)=n(n−1)(n−2)...1=n!$

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:$V_{k}(n)=n⋅n⋅n⋅n...n=n_{k}$

## 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}...k_{m}}(n)=k_{1}!k_{2}!k_{3}!...k_{m}!n! $

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)=(kn )=k!(n−k)!n! $

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}(n)=(kn+k−1 )=k!(n−1)!(n+k−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

- (2 66504

K (2, 8) + K (3, 4) = - Distribution function

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Use the product rule to find out how many three-digit numbers exist. - Examination

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Find the number of all three-digit positive integers that can be put together from digits 1,2,3,4 and which are subject to the same time has the following conditions: on one position is one of the numbers 1,3,4, on the place of hundreds 4 or 2. - Distribution 5016

You have a test with eight questions, where you can choose from 3 answers for each question, and one answer is always correct. The probability that we answer 5 or 6 questions correctly when randomly filling in (that is, we all guess the answers) is ……. Th - Probabilities

If probabilities of A, B, and A ∩ B are P (A) = 0.62, P (B) = 0.78, and P (A ∩ B) = 0.26, calculate the following probability (of the union. intersect and opposite and its combinations): - Probability 80856

The probability of occurrence of a certain phenomenon is the same in all trials and is equal to 0.7. Attempts are repeated until this phenomenon occurs. What is the probability that we will have to make a fifth trial? - First class

The shipment contains 40 items. 36 are first-grade, and four are defective. How many ways can we select five items so that it is no more than one defective? - Bernoulli trial

A used car saleswoman estimates that each time she shows a customer a car, there is a probability of 0.1 that the customer will buy the car. The saleswoman would like to sell at least one car per week. If showing a car is a Bernoulli trial with a probabil - Marbles 3

Paisley has a bag of 16 green, orange, and yellow marbles. If there are eight green marbles, how many marbles have to be yellow for P(yellow) = 1/4? - Ice cream

Annie likes ice cream. In the shop are six kinds of ice cream. How many ways can she buy ice cream in three scoops if each has a different flavor mound and the order of scoops doesn't matter? - Different 68064

Anna painted eggs for art. She had five colors for her eggs. He wants to put three of them on each. How many different colored eggs could she paint? (It's just the colors, not the shapes on them. ) - Four-digit 65124

Please find out how many different four-digit numbers we can create from the digits 3 and 8 so that the two digits three and two digits eight are used in each four-digit number created. - Probability 71784

What is the probability that if you roll the die twice, the sum of 12 will fall? - A fair coin

A fair coin is tossed twice. Write down the set of possible outcomes. What is the probability of obtaining it? I. Exactly one head ii. No head

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