For the permutation the equation is (8!)/(5!) assuming there is no replacement. Which is 8*7*6 = 336
For combinations the equation is (8!)/((5!)(8-5)!) = (8!)/((5!)(3)!) = (8*7*6)/(6) = 56
The number of different permutations of 4 objects taken 4 at a time is calculated using the formula ( n! ), where ( n ) is the number of objects. For 4 objects, this is ( 4! = 4 \times 3 \times 2 \times 1 = 24 ). Therefore, there are 24 different permutations.
The number of permutations of 8 objects taken 2 at a time is calculated using the formula for permutations, which is ( P(n, r) = \frac{n!}{(n-r)!} ). For this case, ( n = 8 ) and ( r = 2 ), so it can be expressed as ( P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56 ). Therefore, there are 56 permutations of 8 objects taken 2 at a time.
There are many formules
Trifecta bets are permutations, not combinations, and there are 6840 of them.
720 permutations (the arrangement matters) 120 combinations (the arrangement doesn't matter)
Do a web search for "permutations and combinations" to find the how. I make it 35,960.
Permutations = 4 x 3 x 2 = 24Combinations = (4 x 3 x 2) / (3 x 2 x 1) = 24/6 = 6
The number of different permutations of 4 objects taken 4 at a time is calculated using the formula ( n! ), where ( n ) is the number of objects. For 4 objects, this is ( 4! = 4 \times 3 \times 2 \times 1 = 24 ). Therefore, there are 24 different permutations.
The number of permutations of 8 objects taken 2 at a time is calculated using the formula for permutations, which is ( P(n, r) = \frac{n!}{(n-r)!} ). For this case, ( n = 8 ) and ( r = 2 ), so it can be expressed as ( P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56 ). Therefore, there are 56 permutations of 8 objects taken 2 at a time.
There are many formules
Trifecta bets are permutations, not combinations, and there are 6840 of them.
720 permutations (the arrangement matters) 120 combinations (the arrangement doesn't matter)
There are 120 permutations and 5 combinations.
8*7/(2*1) = 28
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
There is only one combination. There are many permutations, though.
Combinations from 90 objects, taken 3 at a time is 90C3 = 90*89*88/(3*2*1) = 117,480.