How many combinations of three digits can be made from the numbers 1 - 6 eg 123 431 etc using each digit only once per combination. I'm thinking 60. Just need confirmation?

1. each of the 3 digits can be any one of 6 digits,

so it should be 6 to the third power (108)

2. The above answer is not even close. It makes no sense whatsoever. The answer is 20. See below from Wikipedia

The number of k-combinations (each of size k) from a set S with n elements (size n) is the binomial coefficient (also known as the "choose function"):

n! / (k!(n-k)!)

where n is the number of objects from which you can choose and k is the number to be chosen, and n! denotes the factorial.

As for the answer below (3.): You may be correct in assuming you can use one number repeatedly in one combination (111). The way I read the question is that once you use a digit, it cannot be used again in the same group of 3. Its like asking how many combonations of 3 marbles can be made from 6 distinctly different marbles. One combination could be: Marble A, Marble B, and Marble C. Saying a possible combination could be: Marble A, Marble A, Marble A (same as 111 in the answer below) would be INCORRECT. You don't have 3 marble As just like you don't have three number 1s.

However, even if you can use one number repeatedly in the same group of three, the question asks for the number of combonations you can make, not different numbers you can make. So 121 would be the same as 211. They are different numbers but the same combination of individual digits. Therefore the assumption that the answer is 36 * 6 = 216 is wrong. When you go on and do the numbers starting with 2 you will repeat some combonations. Even more when you start with the 3s and so on.

Answer number 3 correctly answers the question "how many different 3 digit numbers can you make using numbers 1,2,3,4,5, and 6?"

3. I love mathematicians, they solve a problem once and never question it.

Try :

111 112 113 114 115 116

121 122 123 124 125 126

131 132 133 134 135 136

141 142 143 144 145 146

151 152 153 154 155 156

161 162 163 164 165 166

now that's 36 and that's only a sixth of the possibilities, so the answer must be 216.

(or 2 times the original guess)

Wrong, wrong, wrong! This is a simple permutation problem. IF numbers could repeat, it would simply be 6x6x6= 216. HOWEVER, the directions clearly stipulate that numbers cannot repeat. So, on your first selection you have 6 choices; on your second selection you would have 5 choices; and 4 choices on your third selection. So the answer is:

6x5x4 = 120 possible combinations

IF order mattered (it does not here--for instance, here you could pick 123 or 321), then you would divide by the total selection factorial. So, if you are picking from six individuals to be, say, a President, Vice Pres and Treasurer, and the order DOES matter, then the answer would be:

(6x5x4)/(3!) = 20

Or think of it as:

(6x5x4)/(3x2x1)= 5x4 (after cancelling) = 20