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There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
There are 5,461,512 such combinations.
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
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it is hard to say there are lot of combinations belive or not * * * * * If the previous answerer thinks 15 is a lot then true. There are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations. Not so hard to say!
There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
There are 6C3 = 20 such combinations.
There are 5,461,512 such combinations.
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.
In the short form, the total number of different combinations you can get are 64. The work to obtain that number is shown below.Let's call each of them different names. There will be starter 1, 2, 3, 4, and 5.The possible combinations are as follows for the first set of numbers:# 1,2,3,4,5 # 1,2,3,5,4 # 1,2,4,5,3 # 1,2,4,3,5 # 1,3,2,4,5 # 1,3,2,5,4 # 1,3,4,2,5 # 1,3,4,5,2 # 1,4,2,3,5 # 1,4,2,5,3 # 1,4,3,2,5 # 1,4,3,5,2 # 1,5,2,3,4 # 1,5,2,4,3 # 1,5,3,4,2 # 1,5,3,2,4 Those are the possible combinations if the starter 1 went first. Multiply the total combinations for the first set by 4, and you get a total of 64 combinations.
I need to have listed all the 4 number combinations between 1 and 9
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
*inclusive: 74 = 2401 *exclusive: 7*6*5*4 = 840