Elements of combinatorics. Combinatorics

Answer: 3628800 way
Solution.
By the permutation formula, we find:
P (10)= 10! \ u003d 1x2x3x ... x9x10 \u003d 3628800

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3) In how many ways can eight children be seated on eight chairs in the kindergarten's dining room?

SOLUTION

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3) In how many ways can eight children be seated on eight chairs in the kindergarten's dining room?

Answer: 40320 ways
Solution.
By the permutation formula, we find:
P(8) = 8! = 1х2х3х…х7х8=40320

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4) How many different words can be formed by rearranging the letters in the word "triangle" (counting this word itself)?

SOLUTION

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4) How many different words can be formed by rearranging the letters in the word "triangle" (counting this word itself)?

Answer: 39916800 with lov.
Solution.
By the permutation formula, we find:
P (11)= 11! \ u003d 1x2x3x ... x10x11 \u003d 39916800

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5) In how many ways can one person be on duty per day among seven students in a group for 7 days (each must be on duty once)?

SOLUTION

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5) In how many ways can one person be on duty per day among seven students in a group for 7 days (each must be on duty once)?

Answer: 5040 way .
Solution.
By the permutation formula, we find:
P(7) = 7! \u003d 1x2x3x ... x6x7 \u003d 5040
Permutations with repetitions
Any arrangement with repetitions in which element a 1 is repeated k 1 times, element a 2 is repeated k 2 times, and so on. element a n is repeated k n times, where k 1 , k 2 , ..., k n are given numbers, is called a permutation with repetitions of the order