6 over 3, or 6 combination 3, is calculated as (6 x 5 x 4) / (1 x 2 x 3).
3! = 1 x 2 x 3 = 6. The same for any three different digits - you have 6 possible combinations. * * * * * NO! That is the number of permutations! In a combination the order of the numbers does not matter, so there is only one combination 123. And that is the same as 132, or 312 or ...
2.3 recurring
4*3*2*1 = 24 ways.4*3*2*1 = 24 ways.4*3*2*1 = 24 ways.4*3*2*1 = 24 ways.
4!/(2!*2!) = 4*3*2*1/(2*1*2*1) = 6 4!/(2!*2!) = 4*3*2*1/(2*1*2*1) = 6 4!/(2!*2!) = 4*3*2*1/(2*1*2*1) = 6 4!/(2!*2!) = 4*3*2*1/(2*1*2*1) = 6
100=2*2*5*5 so only 1 combination of 3 nos. r der
1/(2/3) = 3/2
Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
3*2 = 6
3 quarters 2 thirds 1 quarter 2 eighths which is the same as 1 quarter
1
There is only 1 combination of the numbers 123. (order does not matter)There are 6 permutations of the numbers 123. (order does matter)1 2 31 3 22 1 32 3 13 1 23 2 1
(2/3) / (1/3) = (2/3) x (3/1) = 2
The answer is 20C4 = 20*19*18*17/(4*3*2*1) = 4845
how many 3/8 ARE IN 1
1 1 1 2 1 3 1 4 2 1 2 2 2 3 2 4 3 1 3 2 3 3 3 4 4 1 4 2 4 3 4 4
1/2 ÷ 3 = 1/2 x 1/3= 1/6