no of words formed by letters of word ENGINEER such that
A) no two vowels are together =240
B) vowels occupy odd positions =48
C) there is exactly one consonant between any two vowels=96
D) G and R both appear before I= 1120
Answers
Given : words formed by letters of word ENGINEER
To Find : Number of Words if :
A) no two vowels are together
B) vowels occupy odd positions
C) there is exactly one consonant between any two vowels
D) G and R both appear before I
Solution:
ENGINEER
8 - Letters
E - 3 Times
N - 2 Times
Vowels = 4 , EEEI
Consonants = G NN R
A) no two vowels are together
Hence CVCVCVCV , VCVCVCVC , VCCVCVCV , VCVCCVCV , VCVCVCCV = 5 ways
4 vowels can be arranged in 4!/3! = 4 Ways
4 Consonants can be arranged in 4!/2! = 12 Ways
Total Ways = 5 x 4 x 12 = 240
vowels occupy odd positions
VCVCVCVC
= 4 x 12 = 48 Ways
there is exactly one consonant between any two vowels
CVCVCVCV , VCVCVCVC - 2 Ways
2 x 4 x 12 = 96
G and R both appear before I
I is 8th then 7!/3!*2! = 420
I is 7th , 8th can be N or E
8th N then 6!/3! = 120
8th E then 6!/2!2! = 180
I is 6th then 7th , 8th can be NE , EN , EE or NN
NE then 5!/2! = 60
EN then 5!/2! = 60
EE then 5!/2! = 60
NN then 5!/3! = 20
I is 5th then 1st 4 can be GR & EE or GR & NN or GR EN
GR & EE => (4!/2! ) * 3!/2! = 36
GR & NN => (4!/2! ) * 3!/3! =12
GR & EN => (4!) * 3!/2! = 72
I is 4th then 1st 3 can be GRE or GRN
GRE => (3! ) * 4!/2!2! = 36
GRN => (3! ) * 4!/3! =24
I is 3rd then
2! * 5!/3!.2! = 20
Adding all = 1120
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