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Re: HOT Competition 1 Sep/8AM: A locker code consists of five digits, for
[#permalink]
02 Sep 2020, 04:52
1
Kudos
Let’s assume that we have five slots. All slots can be filled by 10 digits. So, the number of possible cases with no constraints is 10*10*10*10*10 = 100,000.
We need to exclude cases with 59, such as:
Case 1: 5 9 _ _ _ Case 2: _ 5 9 _ _ Case 3: _ _ 5 9 _ Case 4: _ _ _ 5 9
And:
Case 5: 5 9 5 9 _ Case 6: 5 9 _ 5 9 Case 7: _ 5 9 5 9
Case 1, 2, 3, and 4 each have by 1000 cases, or together 4000 cases. We can see that certain cases are double counted, for example:
- When Case 1 and 3 together result in case 5 such as 5 9 5 9 _ , here _ can be filled with 10 digits, such as 59591, 59592, 59593, 59594,..., 59590, so overall 10 cases here.
- When Case 1 and 4 together result in case 6 such as 5 9 _ 5 9, here _ can be filled with 10 digits, such as 59159, 59259, 59359, 59459,..., 59059, so overall 10 cases here.
- When Case 2 and 4 together result in case 7 such as _5 9 5 9, here _ can be filled with 10 digits, such as 15959, 25959, 35959, 45959,...,0 5959, so overall 10 cases here.
So, these 30 cases are double counted in 4000 cases with 59. Hence, we actually have 3970 cases with 59. 100,000 – 3970 = 96030.
I think that choice C is typo and should be not 96003, but 96030.
Hence C
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Re: HOT Competition 1 Sep/8AM: A locker code consists of five digits, for [#permalink]