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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.

 

 

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

 

A) Vowels are always together.

 

B)vowels are never together.

 

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

 

 

I find such question super confusing….as they involve a few extra steps I guess….i’m grateful if you can take out the time to answer any of these if you know them. You tips and help would is highly appreciated . Thank you!

 

smiley

rosala  Feb 9, 2018
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8+0 Answers

 #2
avatar+82944 
+2

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

 

 

cool cool cool

CPhill  Feb 9, 2018
 #7
avatar+11840 
+1

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???? 

rosala  Feb 10, 2018
 #8
avatar
+1

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.

Guest Feb 10, 2018
 #9
avatar+91797 
+1

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?

Melody  Feb 11, 2018
edited by Melody  Feb 11, 2018
edited by Melody  Feb 11, 2018
edited by Melody  Feb 11, 2018

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