The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
you could make a probability tree if you could be bothered
35
9000
None. You do not have enough numbers to make even one combination.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
you could make a probability tree if you could be bothered
Only one.
Assuming that the six numbers are different, the answer is 15.
35
9000
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
None. You do not have enough numbers to make even one combination.
7
Using the combination fuction, chose three numbers from 45 numbers. The answer is 14,190.
If the 1, 2, 3, 4 and 5 each appear in the numbers (no repeats, no eliminations-- so numbers like 11133 are not valid) then there are 120 combinations. If you can use any of the 5 any number of times up to a max of 5 times, (any possible way of making a 5 digit number using only 1, 2, 3, 4 and 5, so that 33455 for example is valid) then the number of combinations is 3125. The 3125 combinations include the 120 combinations above. This is a permutation of 5 objects, that can be thought of as an arrangement or a rearrangement of the five numbers. The number of permutations of this 5 numbers is equal to 5!. 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Or you can use the graphing calculator to compute 5!, such as: Press 5, MATH, with the right arrow go to PRB, press 4 (for !), ENTER.