Q 8: Count the number of 01-strings with following constraints.
(a) The length is 8. Number of 1s is 2 more than number of 0s.
(b) The length is 8. Number of 1s is 3 more than number of 0s.
(c) The length is 9. Number of 1s is 3 more than number of 0s.
(a) you have 5 1s and 3 0s
so, use the formula for permutations of a set with duplicates:
8!/(5!3!) = 56
That's because for every one of the 8! permutations, the 1s and 0s can be shuffled without altering the string.
(b) and (c) are similar.
Oh, 01-strings, you're so binary-tiful! Let's tackle these constraints one by one, shall we?
(a) The length is 8 and the number of 1s is 2 more than the number of 0s. Interesting! So, let's have 2 0s and 4 1s. Now we need to arrange them in different combinations. Starting with the 0s, we can place them in 8C2 = 28 ways. Next, the 4 1s can be placed in 6C4 = 15 ways. Multiplying these together, we get the total number of 01-strings satisfying this constraint: 28 x 15 = 420.
(b) Now, the length is still 8, but this time there are 3 more 1s than 0s. So, let's have 1 0 and 4 1s. Again, let's arrange them. We have 8C1 = 8 ways to choose the position for 0. Next, we have 7C4 = 35 ways to choose the positions for 1s. Multiplying these, we get 8 x 35 = 280.
(c) Ah, this time the length becomes 9. But don't worry, we can handle it! So, let's have 1 0 and 4 1s, just like before. Now, let's find the positions for them. We have 9C1 = 9 ways to choose the position for 0. Next, we have 8C4 = 70 ways to choose the positions for 1s. Multiplying these, we get 9 x 70 = 630.
So, to summarize:
(a) 420 01-strings with 2 more 1s than 0s and length 8.
(b) 280 01-strings with 3 more 1s than 0s and length 8.
(c) 630 01-strings with 3 more 1s than 0s and length 9.
I hope these answers brought a smile to your binary world!
To count the number of 01-strings with the given constraints, we can use recursion or iterative approaches. In this case, let's use an iterative approach.
(a) The length is 8, and the number of 1s is 2 more than the number of 0s.
In this case, the string can be represented as "01*01*", where '*' denotes any combination of 1s and 0s. We need to figure out the number of ways to insert 1s and 0s in the two '*'.
Let's consider the first '*' position:
- If we place two 0s, then we have 2 remaining positions where we can insert two 1s. So, we have 2C2 = 1 way.
- If we place one 0 and one 1, then we have 2 remaining positions where we can insert one 1 and one 0. So, we have 2C1 * 2C1 = 4 ways.
Therefore, there are a total of 1 + 4 = 5 possible 01-strings with the given constraints.
(b) The length is 8, and the number of 1s is 3 more than the number of 0s.
In this case, the string can be represented as "01*01*01*", where '*' denotes any combination of 1s and 0s. We need to figure out the number of ways to insert 1s and 0s in the three '*'.
Let's consider the first '*' position:
- If we place three 0s, then we have 3 remaining positions where we can insert three 1s. So, we have 3C3 = 1 way.
- If we place one 0 and two 1s, then we have 3 remaining positions where we can insert two 1s and one 0. So, we have 3C1 * 3C2 = 9 ways.
- If we place two 0s and one 1, then we have 3 remaining positions where we can insert one 0 and two 1s. So, we have 3C2 * 3C1 = 9 ways.
Therefore, there are a total of 1 + 9 + 9 = 19 possible 01-strings with the given constraints.
(c) The length is 9, and the number of 1s is 3 more than the number of 0s.
In this case, the string can be represented as "01*01*01*", where '*' denotes any combination of 1s and 0s. We need to figure out the number of ways to insert 1s and 0s in the three '*'.
Let's consider the first '*' position:
- If we place three 0s, then we have 3 remaining positions where we can insert three 1s. So, we have 3C3 = 1 way.
- If we place one 0 and two 1s, then we have 3 remaining positions where we can insert two 1s and one 0. So, we have 3C1 * 3C2 = 9 ways.
- If we place two 0s and one 1, then we have 3 remaining positions where we can insert one 0 and two 1s. So, we have 3C2 * 3C1 = 9 ways.
Therefore, there are a total of 1 + 9 + 9 = 19 possible 01-strings with the given constraints.
To count the number of 01-strings with the given constraints, we need to systematically consider all possible combinations of 0s and 1s that satisfy the given conditions.
(a) The length is 8. Number of 1s is 2 more than the number of 0s.
Let's start by considering the number of 0s. Since the number of 1s is 2 more than the number of 0s, we can represent the number of 0s as x. Therefore, the number of 1s can be represented as x + 2.
Now, we need to find all possible values of x that satisfy the condition. The total number of digits (0s and 1s) is 8. So, the equation becomes:
x + (x + 2) = 8
Simplifying the equation, we get:
2x + 2 = 8
2x = 6
x = 3
Therefore, the number of 0s is 3 and the number of 1s is 5 (since 3 + 2 = 5).
So, the possible 01-strings that satisfy the condition are:
00111000
01011000
01101000
01110000
10011000
10101000
10110000
11001000
11010000
11100000
There are 10 possible 01-strings that satisfy the given condition.
(b) The length is 8. Number of 1s is 3 more than the number of 0s.
Using a similar approach, let's represent the number of 0s as x and the number of 1s as x + 3. The equation becomes:
x + (x + 3) = 8
Simplifying the equation, we get:
2x + 3 = 8
2x = 5
x = 2.5
Since the number of zeros cannot be a fraction, there are no 01-strings that satisfy this condition.
(c) The length is 9. Number of 1s is 3 more than the number of 0s.
Again, representing the number of 0s as x and the number of 1s as x + 3, the equation becomes:
x + (x + 3) = 9
Simplifying the equation, we get:
2x + 3 = 9
2x = 6
x = 3
The number of 0s is 3, and the number of 1s is 6.
So, the possible 01-strings that satisfy the condition are:
001111000
011110000
010111000
101110000
110110000
111010000
There are 6 possible 01-strings that satisfy the given condition.