solve for n where n {N
C(n,2)=45
b. (n+2)!
_____________= 56
n!
To solve for n in the equation C(n, 2) = 45, where C(n, 2) represents the number of combinations of n items taken 2 at a time:
Step 1: Express C(n, 2) using the combination formula:
C(n, 2) = n! / (2!(n-2)!)
Step 2: Substitute C(n, 2) into the equation:
n! / (2!(n-2)!) = 45
Step 3: Simplify the equation:
n! / (2(n-2)(n-3)!) = 45
Step 4: Multiply both sides by 2(n-2)(n-3)! to eliminate the denominator:
n! = 45 * 2(n-2)(n-3)!
Step 5: Simplify the equation further:
n! = 90(n-2)(n-3)!
Step 6: Expand the factorial on the right side:
n * (n-1)! = 90(n-2)(n-3)!
Step 7: Cancel out the (n-1)! on both sides:
n = 90(n-2)(n-3)
Step 8: Expand and simplify the equation:
n = 90(n^2 - 5n + 6)
Step 9: Distribute the 90 on the right side:
n = 90n^2 - 450n + 540
Step 10: Move all terms to one side to get a quadratic equation:
90n^2 - 451n + 540 = 0
At this point, the equation becomes a quadratic equation that can be solved using various methods such as factoring, completing the square, or using the quadratic formula. Once you solve this quadratic equation, you will find the value(s) of n that satisfy the original equation C(n, 2) = 45.
Moving on to the second part of the question:
To solve for n in the equation (n+2)! / n! = 56:
Step 1: Expand the factorials:
(n+2)! = 56n!
Step 2: Write out the factorial as a product:
(n+2) * (n+1) * n! = 56n!
Step 3: Divide both sides by n! to eliminate the common factor:
(n+2) * (n+1) = 56
Step 4: Expand and simplify the equation:
n^2 + 3n + 2 = 56
Step 5: Move all terms to one side to get a quadratic equation:
n^2 + 3n - 54 = 0
Similar to the previous quadratic equation, you can solve this equation using factoring, completing the square, or using the quadratic formula to find the value(s) of n that satisfy the equation (n+2)! / n! = 56.