Permutations and Combinations!!! <3

2.  SQUARE

a) Vowels are always together

Note that  the vowels can appear in any one of these positions  [ V = vowel ]

V V V _ _ _

_ V V V _ _

_ _ V V V _

_ _ _ V V V

There are 4 positions where the vowels can appear together.....and for each of these the vowels can be arranged in 3! ways  =  6 ways    and the other letters can be arranged in 3!  = 6 ways

So......the total possible "words" that can be made where the vowels appear together is

4 *  6 * 6    =   144 "words"  =  144 arangements

b) Vowels are never together 

The vowels can appear in these positions

V _ V _ V _

_ V _ V _ V

So there are 2 possibilities here.....and for each of these, the vowels can be arranged in 3! ways = 6 ways  and the other 3 letters can be arranged in 3! ways  = 6 ways

So.....the total possible arrangements where the vowels never appear together is  just :

2 * 6 * 6    =    72  arrangements

c)  UAE  never appear together

First  note that the total possible  arrangements is just  6!  = 720

Let's count the number of arrangements where UAE  do appear together

U A E _ _ _

_ U A E _ _

_ _ U A E _

_ _ _ U A E

There are 4 of these....and for each, the other letters can be arranged in 3!  = 6 ways

So....the total arrangements where UAE appear together  is just :

4 * 3!  =  4 * 6   =  24

So....the number of arrangements where UAE never appear together  is just

Total arrangements  - Arrangements where UAE appear together  

720  - 24   =  696 arrangements

CPhill i think theres some problem in the 'c' parts answer becoz my teachers answer is 144.

its somewhat like:

No. of ways in which none of U , A , E are together = 3! x ⁴P₃  = 3! x 4!/1! = 3! x 4! = 144.

i have no idea what is done here.

dont you think that U, A , E ..that are vowels...and when they wont appear together, then the answer will be the same as the 'b' part???? 

C)

The 3 letters "UAE" appear in 6 - 3 + 1 = 4 positions. IF these letters are allowed to permute, that is:

UAE, UEA, AEU......etc. Then, you will have: 4 positions x 3! =24 permutations. The remaining 3 letters, S, R, Q will permute in 3! ways =6 ways. So the total permutations will be:

4 x 3! x 3! =4 x 6 x 6 = 144 permutations.

1.Find the no. of permutations of letters a , b , c , d, e , f , g taken all at a time if neither  “beg” nor “cad” patterns appear in any word.

Mmm I'd put a rope around beg 

then that makes a,c,d,f (beg)  so there will be 5!= 120 permutations

If I put a rope around (cad)   that will also be 120 permutaions 

BUT some of these permutations are in both so if I put a rope around (cad) and (beg) that will be 3! =6 permutations.

so altogether the number of permutaions that include the words beg or cad is   120+120-6=  234 permutations.

Without any restrictions there are 7! = 5040 permutations

so the number of permutation without beg or cad is 5040 - 234 = 4806

2.Find the no. of ways in which letters of the word SQUARE can be arranged such that:

A) Vowels are always together.   

SQR  (UAE)   UAE can be arranged in 3!=6 ways and SQR(UAE) can be arranged in 4!=24 ways.

So altogether that is  24*6 = 144 ways

B)vowels are never together.

There are 3 vowels and 3 consonants so the choices are

CVCVCV   or   VCVCVC   

3!*3!*2 = 6*6*2 = 72 ways

C)No. of ways in which none of U , A , E are together. (in any order)

Again, UAE must be in the 1st 3rd and 5th positions   or in the 2nd 4th and 6th positions. I do not think there are any other choices so it is the same as the question above I think.  Maybe I have misunderstood the question?

72 ways

What do you think Rosala, do you question any of my answers (which is perfectly fine) or do you have any questions for me?