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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.

 Jan 12, 2021
 #1
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Stephen forget to include plates with  FIVE  A's     and a plate with SIX A's

 Jan 12, 2021
 #2
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   so    6 C 4  *  26^2    +   6 C 5  * 26    +   6 C 6 = ............ plates with at LEAST four A's.

Guest Jan 12, 2021
 #3
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Please add more thorough explanations I want to understand this.

 Jan 12, 2021
 #4
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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   = .........

Guest Jan 12, 2021
 #5
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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

 

 

cool cool cool

 Jan 12, 2021
 #6
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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

MathSolverForLYFE  Jan 12, 2021
 #7
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AGREED! And good answer CPhill! :D

-Grace

Guest Jan 13, 2021
 #8
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ok. thx

 Jan 15, 2021
edited by Guest  Jan 15, 2021
edited by Guest  Jan 15, 2021
edited by Guest  Jan 15, 2021
edited by Guest  Jan 15, 2021
edited by Guest  Jan 15, 2021
 #9
avatar+112031 
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Why didn't you talk about what you had tried for yourself MathSolver?

There is no evidence that you were trying to solve anything.  So why call yourself a MathSolver?

Looks like you are just asking for other people to solve your homework problem for you.

 Jan 16, 2021
 #10
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Thats HARSH

 Jan 16, 2021
 #11
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It's true though

calclac  Jan 19, 2021

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