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| Math Central | Quandaries & Queries |
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lf I tried I could probably (I think) figure the answer but it would take a minute and I hoped maybe as math people there may be a simple formula… I just watched a new movie Black Phone, the kid is given 5 numbers for a combination lock in order but with no indication where the stops are. The numbers (in order) are 23317, my question is how many possibilities with a standard combination lock are there…ie. 2-33-17, 23-3-17, 23-31-7…. As you can imagine not a great movie…my |
Hi Kendal.
You have 5 digits in a row and hence 4 places where you can insert the stops. You get to choose 2 of the 4 spaces to insert them. Thus there are "4 choose 2" ways to place the stops and hence "4 choose 2" possible combinations. If you know the formula for "4 choose 2" you can calculate the answer. If you just want the answer type 4 choose 2 into you web browser. But this is no fun.
You have 5 digits in a row and hence 4 places where you can insert the stops. Here is how I see it, the letter b stands for blank space.
2b3b3b1b7
You get too choose two of the b and change them to s which stands for stop. Start by choosing one b and change it to an s. There are 4 choices. Here is one of them
2b3b3s1b7
Regardless of which b you chose there are three more b and you get to convert one of them to an s. Here is one of them.
2s3b3s1b7
Thus there are $4 \times 3 = 12$ possibilities. However you could have arrived at
2s3b3s1b7
by placing the s between the 2 and 3 first and the s between the 3 and 1 second. Thus each of the possibilities appears twice. Hence there are only $12/2 = 6$ possible lock combinations.
Penny
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