ow many binary strings of length 8 are there that contain two or fewer 1s?
Question
ow many binary strings of length 8 are there that contain two or fewer 1s?
Solution
To solve this problem, we need to consider three cases: strings with no 1s, strings with one 1, and strings with two 1s.
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Strings with no 1s: There is only one such string, which is 00000000.
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Strings with one 1: There are 8 such strings. The 1 can be in any of the 8 positions.
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Strings with two 1s: We need to choose 2 positions out of 8 to place the 1s. This is a combination problem, and can be solved using the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. So, C(8, 2) = 8! / [2!(8-2)!] = 28.
Adding up these three cases, we get 1 + 8 + 28 = 37. So, there are 37 binary strings of length 8 that contain two or fewer 1s.
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