SOLUTION: Using the letters of the word YOUNG, tell how many different 5-letter combinations are possible if: a) The first letter must be Y b) The vowels and consonants alternate,begin

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Question 358137: Using the letters of the word YOUNG, tell how many different 5-letter combinations are possible if:
a) The first letter must be Y
b) The vowels and consonants alternate,beginning with a consonant (Y is a consonanthere and Y does not have to be first)
Please help im lost with this question i would appreciate it

Found 3 solutions by jrfrunner, Edwin McCravy, sudhanshu_kmr:
Answer by jrfrunner(365)   (Show Source): You can put this solution on YOUR website!
a) The first letter must be Y
Since there only 4 letters free to be chosen, the answer = 4!=24
==
b) The vowels and consonants alternate,beginning with a consonant
-
let C=consonant, V=vowel
There are 3 C and 2 V
CVCVC
Number of ways 3*2*2*1*1=12

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Using the letters of the word YOUNG, tell how many different 5-letter combinations are possible if:
a) The first letter must be Y
Y _ _ _ _

There is 1 way to choose the 1st letter, since it must be a Y.
There are 4 letters left to choose for the 2nd letter.
There will then be 3 letters left to choose for the 3rd letter.
There will then be 2 letters left to choose for the 4th letter.
There will then be only 1 letter left to choose the for 5th letter.

Answer 1*4*3*2*1 = 24.

Checking:

 1.  YOUNG
 2.  YOUGN
 3.  YONUG
 4.  YONGU
 5.  YOGUN
 6.  YOGNU
 7.  YUONG
 8.  YUOGN
 9.  YUNOG
10.  YUNGO
11.  YUGON
12.  YUGNO
13.  YNOUG
14.  YNOGU
15.  YNUOG
16.  YNUGO
17.  YNGOU
18.  YNGUO
19.  YGOUN
20.  YGONU
21.  YGUON
22.  YGUNO
23.  YGNOU
24.  YGNUO

b) The vowels and consonants alternate,beginning with a consonant (Y is a consonant there and Y does not have to be first)
They can only alternate this way:

consonate, vowel, consonant, vowel, consonant

There are 3 consonants to choose for the 1st letter.
There are 2 vowels to choose for the 2nd letter.
There will then be 2 consonants left to choose for the 3rd letter.
There will then be only 1 vowel left to choose for the 4th letter.
There will then be only 1 consonant left to choose for the 5th letter.

That's 3*2*2*1*1 = 12 ways

Checking:

 1.  YONUG
 2.  YUNOG
 3.  YOGUN
 4.  YUGON
 5.  NOYUG
 6.  NUYOG
 7.  NOGUY
 8.  NUGOY
 9.  GONUY
10.  GUNOY
11.  GOYUN
12.  GUYON

Edwin

Answer by sudhanshu_kmr(1152)   (Show Source): You can put this solution on YOUR website!

Here 3 consonants and 2 vowels.
A) no. of words where first letter is y = 4P4 = 4! = 24
( no. of ways to arrange 4 letters O,U,N and G on four places i.e YXXXX )



............................................................................
B)
now, words must be in format CVCVC

where c- consonants, v -vowels
For first place we can put either N or G, because Y can't be on first place.
no. of ways = 2

For second place we can put either O or U. so, no. of ways = 2
similarly, for third place, no. of ways = 2 ( Y or remaining of N , G )
For fourth place , no. of ways = 1 (only one vowel left)
For fifth place, no. of ways = 1 (only one consonants left)

so, no. of words = 2* 2 * 2 * 1 * 1 = 8



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