+28 Ms. Q gave Grogg the following problem: "License plates in Aopslandia consist of six upper-case letters. For example, two possible Aopslandian license plates are ABCDEF and AAAOPS. No two license plates are the same. How many possible Aopslandian license plates are there which contain exactly four A's, or exactly two B's, or both?" Grogg got the answer (64)⋅262+(62)⋅264, but Ms. Q told him this was the wrong answer! (a) How did Grogg arrive at his answer? (b) Why is Grogg's answer wrong? Should the correct answer be smaller or larger than Grogg's answer (and why)? (c) Write a solution to Ms. Q's problem, explaining in complete sentences what the correct answer to the problem should be and why.
Note: This is a different problem than
https://web2.0calc.com/questions/ep-melody-cphil-please-come
Thank you
+750 I haven't figured out a or b, but here is the answer to part c:
1. Find the number of license plates with exactly four A's:
Choose 4 positions out of 6 for the A's: 6C4 = 15 ways
Fill the remaining 2 positions with any of the 26 letters: 26 * 26 = 676 ways
Total: 15 * 676 = 10,140 ways
2. Find the number of license plates with exactly two B's:
Choose 2 positions out of 6 for the B's: 6C2 = 15 ways
Fill the remaining 4 positions with any of the 26 letters: 26 * 26 * 26 * 26 = 456,976 ways
Total: 15 * 456,976 = 6,854,640 ways
3. Find the number of license plates with both four A's and two B's:
Choose 4 positions for the A's: 6C4 = 15 ways
Choose 2 of the remaining 2 positions for the B's: 2C2 = 1 way
Fill the remaining 0 positions: 1 way
Total: 15 * 1 * 1 = 15 ways
4. Find the total number of license plates:
Total = (Number with 4 A's) + (Number with 2 B's) - (Number with both)
Total = 10,140 + 6,854,640 - 15
Total = 6,864,765 ways
Therefore, there are 6,864,765 possible Aopslandian license plates that contain exactly four A's, or exactly two B's, or both.
+130458 Here's my best attempt
4 A's
Choose any 4 of 6 positions for the A's = C (6,4)
For the other two positions we can choose 1 of 25 remaining letters for one position and then 1 of any 24 remaining letters (We can't choose 2 B's )
So
C(6,4) * 25 * 24 = 9000
2 B's
Choose any 2 of 6 positions for the B's
For the other positions we can choose any 1 of 25 letters for 1 position and (24)^3 for the other 3 positions ( we can't have 2 A's)
C(6,2) * 25 * 24^3 = 5,184,000
4A;s, 2B's = All the arrangements for the "word" AAAABB = 6! / ( 4! * 2!) = 15
I get
9000 + 5,184,000 + 9000 = 5,202,000 plates
