How many ways can you scramble the letters.?

You have double letters (C, A, L) and then the rest (U, T, O, R) are single letters

What you could do is consider 3 cases:
1) The string has no pairs of repeated letters
2) The string has one pair of repeated letters
3) The string has two pairs of repeated letters

CASE 1 - Zero pairs of repeated letters

You have 7 unique letters and you want to permute 4 of them:
7P4 = 7! / (7-4)! 
= 7 * 6 * 5 * 4
= 840 ways

CASE 2 - One pair of repeated letters

First choose which letter is repeated:
3C1 = 3 ways
Then from the remaining *6* unique letters, pick 2.
6C2 = 15 ways
Now take your 4 chosen letters, with 2 repeats and arrange them:
4! / (2! 1! 1!) = 12 ways

Multiply these all together:
3 * 15 * 12
= 540 ways

CASE 3 - Two pairs of repeated letters

First, from the 3 doubled letters, choose 2:
3C2 = 3 ways
Now take your 4 chosen letters, with 2 *pairs* of repeats and arrange them:
4! / (2! 2!) = 6 ways

Multiply these together:
3 * 6
= 18 ways

ADD THEM UP:
Case 1 = 840
Case 2 = 540
Case 3 = 18

Total = 1398

If you were familiar with generating functions, that's another way to solve this. Feel free to look up more about that if you are so inclined.

Here's a sneak preview of this method:
4 single letters --> (1 + x)^4
3 double letters --> (1 + x + x²/2!)^3

Multiplying that out:
(1 + x)^4 * (1 + x + x²/2!)^3
= (1/8)x^10 + (5/4)x^9 + 6x^8 + 18x^7 + (297/8)x^6 + (219/4)x^5 + (233/4)x^4 + 44x^3 + (45/2)x^2 + 7x + 1

Take the coefficient of x^4 (since we are using a 4 letter subset).
(233/4) and rewrite it as n/4!
= (6*233)/(6*4)
= 1398 / 4!
n = 1398

Answer:
1398 ways