What are the values in the degree 7 row of Pascal’s triangle?
The values in the degree 7 row of Pascal's triangle are 1, 7, 21, 35, 35, 21, 7, 1.
To find the values in the degree 7 row of Pascal's triangle, we can use the formula for the binomial coefficients. In Pascal's triangle, each number is the sum of the two numbers directly above it.
To find the values in the degree 7 row, we start with the first number, which is always 1. Then we can use the formula to find the remaining numbers. The formula for the kth number in the nth row is:
C(n, k) = n! / (k! * (n - k)!)
Let's calculate the numbers in the degree 7 row step-by-step:
C(7, 0) = 7! / (0! * (7 - 0)!)
= 1
C(7, 1) = 7! / (1! * (7 - 1)!)
= 7
C(7, 2) = 7! / (2! * (7 - 2)!)
= 21
C(7, 3) = 7! / (3! * (7 - 3)!)
= 35
C(7, 4) = 7! / (4! * (7 - 4)!)
= 35
C(7, 5) = 7! / (5! * (7 - 5)!)
= 21
C(7, 6) = 7! / (6! * (7 - 6)!)
= 7
C(7, 7) = 7! / (7! * (7 - 7)!)
= 1
Therefore, the values in the degree 7 row of Pascal's triangle are: 1, 7, 21, 35, 35, 21, 7, 1.
To find the values in the degree 7 row of Pascal's triangle, we need to understand the pattern of the triangle. Pascal's triangle is a triangular array of numbers in which each number is the sum of the two numbers directly above it.
Here is an example of Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
To get the values in the degree 7 row, you need to know a few things:
1. The first row of Pascal's triangle is always [1].
2. Each row starts and ends with 1.
3. To get the numbers in between, you add the two numbers directly above it.
Now, let's find the values in the degree 7 row:
The degree 7 row will have 8 elements since we start counting from 0. The row will look like this:
1 _ _ _ _ _ _ _ 1
To fill in the missing values, we can use the formula for calculating each element:
C(n, r) = n! / (r!(n-r)!),
where C(n, r) represents the combination of n objects taken r at a time.
For the degree 7 row, n = 7 and r will vary from 0 to 7.
Using the formula, we can calculate the missing values:
C(7, 0) = 7! / (0!(7-0)!) = 1
C(7, 1) = 7! / (1!(7-1)!) = 7
C(7, 2) = 7! / (2!(7-2)!) = 21
C(7, 3) = 7! / (3!(7-3)!) = 35
C(7, 4) = 7! / (4!(7-4)!) = 35
C(7, 5) = 7! / (5!(7-5)!) = 21
C(7, 6) = 7! / (6!(7-6)!) = 7
C(7, 7) = 7! / (7!(7-7)!) = 1
So, the degree 7 row of Pascal's triangle is:
1 7 21 35 35 21 7 1