Question 335824: The correct combination of a lock is first turn right to d1, then turn left to d2, and last turns to d3, where d1, d2, and d3 are selected from 0, 1, 2, ..., 50. How many different lock combinations are possible with such a lock, if
a) the numbers may be repeated?
b) the numbers may not be repeated?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! On your first turn, you may pick 50 numbers.
On your second turn, you may pick 50 numbers, if they can be repeated, and you may pick 49 numbers, if they cannot be repeated.
On your third turn, you may pick 50 numbers, if they can be repeated, and you may pick 48 numbers, if they cannot be repeated.
If the numbers can be repeated, your total possible combinations may be:
50 * 50 * 50.
If the number cannot be repeated, your total possible combinations may be:
50 * 49 * 48.
To see this more clearly, assume the numbers are from 1 to 3 only.
With repetition, you can have 3 * 3 * 3 = 27 possible combinations.
Without repetition, you can have 3 * 2 * 1 = 6 possible combinations.
Assuming numbers 1 to 3 only, here are the total possible combinations with repetition.
111 112 113
121 122 123
131 132 133
211 212 213
221 222 223
231 232 233
311 312 313
321 322 323
331 332 333
That's a total of 9 * 3 = 27 possible combinations.
Here are the total possible combinations without repetition.
123
132
213
231
312
321
With 50 numbers, the number of possible combinations becomes much larger, but the concept remains the same.
With 50 possible numbers and repetition allowed, you got:
50 * 50 * 50
With 3 possible numbers and repetition allowed, you got:
3 * 3 * 3
With 50 possible numbers and repetition not allowed, you got:
50 * 49 * 48
With 3 possible numbers and repetition not allowed, you got:
3 * 2 * 1
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