+71 Ms. Q gave Steven the following problem: "License plates in Acardia consist of six upper-case letters. For example, two possible Acardian license plates are $ABCDEF$ and $AAAOPS$ No two license plates are the same. How many possible Arcadian license plates are there which contain at least four $A$'s?"
Steven$\binom{6}{4} \cdot 26^2,$ got the answer but Ms. Q told him this was the wrong answer!
(a) How did Steven arrive at his answer?
(b) Why is Steven's answer wrong? Should the correct answer be smaller or larger than Steven'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.
+71 Please add more thorough explanations I want to understand this.
there is 6 C 4 ways to place 4 A's on the plate leaving two positions where you can place 26 letters
6 C 4 * 26 * 26
there is 6 C 5 ways to place 5 A's (that is SIX ways) with one more position to place one of 26 letters
6 * 26
finally there is ONE way to have SIX A's
6
6 C 4 = 15 soo 15 * 26 * 26 + 6* 26 + 1 = .........
+129847 There is only one plate that contains 6 As
Five As
Choose any 5 of 6 positions for the As
Then....for the other position any one of the other 25 letters can be chosen =
C(6,5) * C(25,1) = 6*25 = 150
4 As
Choose any 4 of 6 positions for the As = C(6,4) = 15
And for each of these, we have two possibilities
(a)The same letter can occupy the other two positions = 25 possibilities
(b)Two different letters can be chosen for the other two positions = 25 * 24
And we can arrange these in 2 ways
So...the total number of plates possible =
6As + 5As + 4As =
1 + 150 + 15 ( 25 + 25*24*2) =
1 + 150 + 15 ( 25 + 1200) =
1 + 150 + 15 ( 1225) =
1 + 150 + 18375 =
18526 possible plates
+71 Yoooooo! Thanks a lot! I'll read over it again. Have a great day and let's hope that 2021 brings cheer instead of more fear :-D
ok. thx
