number of arrangements without restriction = 9!/2 = 181440 (There were 2 N's, so we divide by 2!)
We have 3 vowels, we don't want any 2 to be beside each other.
so consider the possible pairs.
IU IA UA UI AI AU , that is there are 6 of the, or P(3,2)
So if we consider a pair of vowels to be X , the number of ways for the vowels to be together is
8! (6)/2 = 120960
(notice that would include the cases where the 3 vowels are together
e.g. IAHNDUSTN
and IAUHNDSTN
number of ways for the vowels to be apart
= 181440 - 120960 = 60480
b) The relative position of vowels and consonants must remain the same, so we need
CVCCVCCVC , given 6 cosonants (2 the same) and 3 vowels
or 6x3x5x4x2x3x2x1x1/2! = 2160
check my arithmetic
From the word HINDUSTAN =====
(a) In how many of these anagrams none of the vowels come together.
(b) In how many of these anagrams do the vowels and the consonants occupy the same relative positions as in HINDUSTAN?
1 answer
1 answer