Compute the number of eleven-digit integers that can be formed using all of seven 1s and four 0s if you never start or end with a 0 and no 0s are ever next to each other.
Start with seven 1's next to each other.
Add four zeroes to the six spaces between the ones.
There are C(6,4) ways to do this.
Hence the same number of 11 digit numbers.
Can you explain the process of finding the answer more I'm a little confused
To compute the number of eleven-digit integers that can be formed using all seven 1s and four 0s, while satisfying the given conditions, we can break the problem into smaller steps:
Step 1: Determine the possible positions for the zeros.
Since zeros cannot be next to each other, we need to distribute the four zeros among the available positions for the digits. There are a total of eleven positions (digits) in the integer, and there are seven 1s. Thus, we can place one zero between each pair of 1s, as well as one zero on either side of the string of 1s.
_1_1_1_1_1_1_1_
To distribute the four zeros among the seven available positions, we can use combinations. We want to choose four positions for the zeros from the seven available positions:
C(7, 4) = 7! / (4!(7 - 4)!) = 7! / (4!3!) = (7 × 6 × 5) / (3 × 2 × 1) = 35
So, there are 35 possible ways to place the zeros in the eleven-digit integer following the given conditions.
Step 2: Determine the possible positions for the ones.
Since we know the positions of the zeros, we can determine the remaining positions for the ones. Each remaining position can be filled with a one.
In the example below, the underscores represent possible positions for the ones:
_101_1_1_1_1_1_
To determine the number of possible positions for ones:
- The first underscore on the left represents the position before the first zero and has two possibilities (0 or 1).
- The second underscore represents the position between the first two zeros and also has two possibilities.
- The third underscore represents the position between the next two zeros and has two possibilities as well.
There are four such positions, so the total number of possible positions for the ones is:
2 × 2 × 2 × 2 = 2^4 = 16
So, there are 16 possible positions for the ones.
Step 3: Compute the total number of possible combinations.
Finally, we can multiply the number of possible positions for the ones by the number of possible positions for the zeros:
35 × 16 = 560
Thus, there are 560 eleven-digit integers that can be formed using all seven 1s and four 0s, while also satisfying the given conditions.