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How many 3-letter words can we make from the letters A, B, C, and D, if we are allowed to repeat letters, and we must use the letter A at least once, and the letter B at least once?

 
 Jun 5, 2021
 #1
avatar+122 
+2

I believe that the answer is 18.  

 

Counting is a difficult mathematical topic for me, so make sure to check my work.  

 

With repeats being tricky to count, I honestly think it is easier to simply calculate it manually, as opposed to with complicated counting tricks.  

 

1.  Repeats A twice.  

 

There are 3 combinations, revolving around where to put the B

 

BAA

ABA

AAB

 

2.  Repeats B twice.  

3 combinations, revolving around where to put the A

 

ABB

BAB

BBA

 

3.  Does not repeat A or B.  

Because it is impossible to repeat a C, or a D, repeats are nonexistent here.  

First, we place the A, then B, resulting in 6 total combinations.  

 

After that, we have one space left for either a C or a D, making the total for this case as 6*2 or 12.  

 

ABc  ABd

 

BAc  BAd

 

AcB  AdB

 

BcA  BdA

 

cAB  dAB

 

cBA   dBA

 

 

In total, there are 12+3+3 combinations or 18 total.  

 
 Jun 5, 2021
 #2
avatar+119772 
+1

A, B  appear once  

Choose  any 2 of 3 positions for them  =  C(3,2)  = 3   and arrange  each of these in 2 ways

And  for each of these arrangements, we have 2 choices for the remaining letter   (C or D)

So   3 * 2 * 2   =   12 words

 

ABX     AXB    XAB

BAX     BXA    XBA     

 (2 choices   for X  in each case)

 

A or B appear twice

Choose  2 of the 3 positions  for either letter = 2*C(3,2)  =  6

And for each of these....we  have  1 choice   (only B  or A)   

So  6 * 1   =   6  words

 

AAX     AXA     XAA

BBX     BXB     XBB

(1 choice for X in ieach case)

 

I get      12  +  6   =  18  "words"   (same  as  EnchantedLava  )

 

 

cool cool cool

 
 Jun 5, 2021

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