![]() Write the smallest and largest 2-digit natural number. The first student will solve a similar problem with a probability of 0.6, the second student will solve at a probability of 0.55, and the third will solve at a probability of 0.04. Three students independently try to solve the problem. A) How many such numbers are there? You solve using a tree diagram. How many ways can this be done?Ĭreate all four-digit numbers in which the digits 0, 2, 5, and 9 do not repeat. The group of 10 girls should be divided into two groups with at least four girls in each group. ![]() How many ways can five letters be arranged? How many different ways can you arrange your reindeer?įrom how many elements can we create 990 combinations, 2nd class, without repeating? You always have your reindeer fly in a single-file line. You have four reindeer, and you want to have 3 fly your sleigh. What is the probability that someone will guess our PIN code on the first try? We used the digits 2, 3, 4, 5, and 7 when entering the PIN code, and we only used each digit once. In what order to fall goals? How many game sequences were possible during the game? How many options are there that can squirt? ![]() How many options if no two girls sit next to each other?Ģ0 colored balls in a bag: Four red Seven green Nine yellow What is the probability of picking a yellow ball? There are five girls and seven boys in the group. How many options do we have?įoundation of combinatorics in word problems k is logically greater than n (otherwise, we would get ordinary combinations).Ĭ k ′ ( n ) = ( k n + k − 1 ) = k ! ( n − 1 ) ! ( n + k − 1 ) ! Įxplanation 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.Ī typical example is: we go to the store to buy 6 chocolates. Here we select k element groups from n elements, regardless of the order, and the elements can be repeated. Their number is a combination number and is calculated as follows:Ĭ k ( n ) = ( k n ) = k ! ( n − k ) ! n ! Ī typical example of combinations is that we have 15 students and we have to choose three. In mathematics, disordered groups are called sets and subsets. The elements are not repeated, and it does not matter the order of the group's elements. k m ! n ! Ī typical example is to find out how many seven-digit numbers formed from the numbers 2,2,2, 6,6,6,6.Ī combination of a k-th class of n elements is an unordered k-element group formed from a set of n elements. Repeating some (or all in a group) reduces the number of such repeating permutations. n = n k Permutations with repeatĪ repeating permutation is an arranged k-element group of n-elements, with some elements repeating in a group. We calculate their number according to the combinatorial rule of the product: A typical example is the formation of numbers from the numbers 2,3,4,5, and finding their number. 1 = n !Ī typical example is: We have 4 books, and in how many ways can we arrange them side by side on a shelf?Ī 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. The elements are not repeated and depend on the order of the elements in the group. It is thus any n-element ordered group formed of n-elements. The permutation is a synonymous name for a variation of the nth class of n-elements. For calculations, it is fully sufficient to use the procedure resulting from the combinatorial rule of product. The notation with the factorial is only clearer, equivalent. N! we call the factorial of the number n, which is the product of the first n natural numbers. 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. The number of variations can be easily calculated using the combinatorial rule of product. The elements are not repeated and depend on the order of the group's elements (therefore arranged). C k ( n ) = ( k n ) = k ! ( n − k ) ! n ! n = 1 1 k = 3 C 3 ( 1 1 ) = ( 3 1 1 ) = 3 ! ( 1 1 − 3 ) ! 1 1 ! = 3 ⋅ 2 ⋅ 1 1 1 ⋅ 1 0 ⋅ 9 = 1 6 5 Number of combinations: 165Ī bit of theory - foundation of combinatorics VariationsĪ variation of the k-th class of n elements is an ordered k-element group formed from a set of n elements.
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