Compute the number of eleven-digit integers that can be formed using all of seven 1s and four 0s if you never start of end with a 0 and no 0s are ever next to each other.
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To compute the number of eleven-digit integers that can be formed using seven 1s and four 0s with the given restrictions, we can break it down into steps:
Step 1: Place the four 0s in the eleven-digit integer.
Since no 0s can be next to each other, we need to determine the positions where we can place the 0s. We have eleven positions available, and we need to choose four of them for the 0s. This can be done using the combination formula.
The number of ways to choose four positions out of eleven is given by the combination formula: C(n, r) = n! / (r!(n-r)!).
In this case, n = 11 (number of available positions) and r = 4 (number of 0s to be placed). So, the number of ways to place the 0s is:
C(11, 4) = 11! / (4!(11-4)!) = 330
Step 2: Place the seven 1s in the remaining positions.
After placing the 0s, we are left with seven positions to place the seven 1s. This can be done in a straightforward manner since there are no restrictions for the placement of the 1s.
Therefore, the number of ways to place the 1s is simply 7!.
Step 3: Multiply the number of ways from Step 1 and Step 2.
To obtain the total number of eleven-digit integers, we need to multiply the number of ways to place the 0s (Step 1) and the number of ways to place the 1s (Step 2) since these steps are independent.
Total number of eleven-digit integers = (number of ways to place 0s) x (number of ways to place 1s)
= 330 x 7!
Hence, the total number of eleven-digit integers that can be formed using all of seven 1s and four 0s with the given restrictions is 168,120.