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nicolesan3
26.04.2021 •
Mathematics
Why is 6P4 = 360 but 6C4 = 15?
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Ответ:
Short answer (I write nPr = P(n, r) and nCr = C(n, r) ):
P (6, 4) = 6! / (6 - 4)! = 6! / 2! = 720 / 2 = 360
C (6, 4) = P (6, 4) / 4! = 6! / (4! (6 - 4)!) = 360 / 24 = 15
Long
P(n, r) counts the number of permutations of n objects taken r at a time, given by
P(n, r) = n ! / (n - r )!
A permutation is a unique arrangement of objects such that the order in which they are arranged is taken into account. For example, if the objects in question are the numbers in the set {1, 2, 3}, then
• there are 3! = 6 total possible permutations if we take all 3 numbers at once:
123, 132, 213, 231, 312, 321
• there are 3!/(3-2)! = 3!/1! = 6 total permutations if we only take 2 numbers at once:
12, 13, 21, 23, 31, 32
• there are 3!/(3-1)! = 3!/2! = 3 total permutations if we take only 1 number at a time:
1, 2, 3
• and there is 3!/(3-0)! = 3!/3! = 1 way of permuting the 3 numbers without taking any of them:
(the permutation itself is just empty space)
By contrast, C(n, r) counts the combinations of n items taken r at a time, given by
C(n, r) = P(n, r) / r !
A combination is like a permutation, but the order of the objects doesn't matter. Continuing with the previous example of arrangements of the numbers from {1, 2, 3}, we have
• 3! / (3! (3-3)!) = 1 combination taking all 3 numbers at once:
123
(the other 5 permutations listed earlier are made up of the same numbers, so we consider them duplicates)
• 3! / (2! (3-2)!) = 3 combinations taking only 2 numbers at once:
12, 13, 23
• 3! / (1! (3-1)!) = 3 combinations taking only 1 number:
1, 2, 3
• 3! / (0! (3-0)!) = 1 combination taking none of them:
(again, empty space)
The main point is that the order of objects is considered across permutations, while it's ignored across combinations.
Ответ:
The inside of a saucepan has a cylindrical shape so the capacity of a saucepan is:![V=\pi .(\frac{20}{2})^{2}.13=4084(cm^{3})=4.084(l)](/tpl/images/0908/5248/e4375.png)
Step-by-step explanation: