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4 -letter "words" are formed using the letters A, B, C, D, E, F, G. How many such words are possible for each of the following conditions?

 

a) No condition is imposed.

 

b) No letter can be repeated in a word.

 

c) Each word must begin with the letter A.

 

d) The letter C must be at the end.

 

e) The second letter must be a vowel.

 

So my class went over this today and I couldn't understand what to do for each. We're supposed to use either 

 

nPr

Factorial

nCr

(and another I can't remember the name of)

 

to solve for the answer.

 

I just need someone to explain this to me! Thank you!cool

jrod0116  Nov 9, 2017
 #1
avatar+86948 
+1

a) No condition is imposed

We can use any of the seven letters in each position...so

7 * 7 * 7 * 7  =  2401 "words"

 

b) No letter can be repeated in a word.

We have 7 ways to pick the first letter, 6 ways to pick the second, 5 ways to pick the third and 4 ways to pick the fourth............so   7 * 6 * 5 * 4  =  840 "words"

 

c) Each word must begin with the letter A

If we can repeat letters, the other three positions can be filled in 7 * 7 * 7 =  343 ways....so..we have 343 "words"

If we cannot repeat letters we have   6 * 5 * 4  ways to fill the other three positions = 120 "words"

 

d) The letter C must be at the end.

Same answers as  (c)

 

e) The second letter must be a vowel.

If we can repeat letters....we have 7 ways to fill three of the positions and 2 ways to fill the remaining one [ since we have only two vowels ]...so....   7 * 2 * 7 * 7  = 686 "words"

If we cannot repeat letters, we have 7 * 2 * 6 * 5  =  420 "words"

 

 

cool cool cool

CPhill  Nov 9, 2017
 #2
avatar+8 
+2

Thank you, CPhill, this was really helpful. I especially liked that you told me your logic... very cool.

 

Jrod

jrod0116  Nov 10, 2017

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