Question

How do I find n in the the permutation equation of nP3 = 990

Answer 1

To find n in the permutation equation nP3 = 990, expand the factorial expression and solve the simplified equation n * (n - 1) * (n - 2) = 990. By factorizing 990, we find that n = 11.

Step-by-step explanation:

To find n in the permutation equation nP3 = 990, we need to understand that the permutation formula for selecting 3 items out of n is given by nP3 = n! / (n - 3)!. Since nP3 = 990, we need to solve for n in the equation n! / (n - 3)! = 990.

Starting with the factorial expressions, we can expand the numerator until we reach the (n-3)! term so we can cancel it out with the denominator:

• n! = n * (n - 1) * (n - 2) * (n - 3)!
• (n - 3)! cancels out in both the numerator and the denominator.
• Thus, n * (n - 1) * (n - 2) = 990

We then look for a combination of three consecutive numbers that, when multiplied together, equal 990. By trial and error or by using prime factorization to break down 990, we find that it can be expressed as 10 * 9 * 11. Therefore, n must be 11, because 11 is the largest number in the consecutive sequence.

Answer 2

n=11

Step-by-step explanation:

Permutations

The formula to calculate permutations is:

`\displaystyle nPm=(n!)/((n-m)!)`

We know nP3=990. Substituting:

`\displaystyle nP3=(n!)/((n-3)!)=990`

Expanding the factorial in the numerator:

`\displaystyle (n(n-1)(n-2)(n-3)!)/((n-3)!)=990`

Simplifying:

`n(n-1)(n-2)=990`

This equation can be easily solved by inspection. The product of three consecutive numbers resulting 990. Starting from greatest to smallest, those numbers are 11,10, and 9, thus:

`n(n-1)(n-2)=11*10*9`

It follows that n=11