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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!
#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 whereUAE 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
#7
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????
#8
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.
#9
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?