Really Amazing Dudes

Main idea : Fix RAD at the front and compute the number of permutations of the remaining letters.

Complete proof :

Call "RAD!!!"-ness points Rp for short; not to be confused with Rupiah .

Since all permutations considered for Rp begin with RAD, we can fix RAD at the front and ignore them. This way, RADLEIGH becomes LEIGH, which has $5! = 120$ permutations, and BRADLEY becomes BLEY, which has $4! = 24$ permutations. Thus Radleigh's Rp is $120 + (120 - 24) = \boxed{216}$ .

Motivation : Eh, it's pretty obvious you want to compute the number of permutations of each name.

Generalization : Using different names is not fun. Even when they have multiple R, A, D, as they are indistinguishable, they have no effect; we can still pick an arbitrary R, A, and D to remove and arrange the rest using multinomial coefficients. A more fun generalization is "!!RAD!!"-ness points, where RAD can occur anywhere in the permutation. With at most one R, at most one A, or at most one D, this is easy: gather the RAD into a single block and use multinomial coefficient again. When there are two or more of each letter though (like ARDRAD), it becomes pretty hard, as RADRAD should be counted only once (and not twice due to the double RAD blocks).

Radleigh's name must start with "RAD", but the other $5$ letters can be in any order. Because all $5$ of the letters are distinct, there are $5!=\textbf{120}$ ways to arrange Radleigh's name so it starts with "RAD". For Bradley, his name must start with "RAD", but the other $4$ letters can be in any order. Because all $4$ of the letters are distinct, there are $4!=\textbf{24}$ ways to arrange Bradley's name so it starts with "RAD".

The difference between these is $96$ . Based off of Sameer's metric and Adam's information, Radleigh has $120+96=\boxed{216}$ "RAD!!!"-ness points.

Consider "RADleigh" and "bRADley".

• RADleigh: RAD* * * * * * * * * -> 5 missing letters to choose. ( $5!$ ways)
• bRADley: RAD* * * * * * -> 4 missing letters to choose. ( $4!$ ways)

Difference $= 5! -4! = 96$ .

If Bradley has 120 (less) "RAD!!!"-ness points, Radleigh will have $120+96 = \boxed{216}$ points.

PS: This question almost confused me that number of ways = "RAD!!!"-ness points.

First, we calculate that Bradley has $4!=24$ permutations of his name starting with "Rad". Since the points have a one to one ratio to the permutations, they must equal $120-24=96$ more than the actual permutations. We thus deduce that Radleigh has $5!+96=120+96=\boxed{216}$ "Rad"-ness points.

Why such a high rating?

Radleigh has 5! - 4! = 96 more points than Bradley, so he has 120+96=216... Trivial