Math, asked by maggie42, 7 months ago

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

Answered by amitnrw
3

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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