Permutaion

first of all, condition 'two consonants remain together' comes as ONE
in total, there are 4 consonants (R,C,T,N); when they come 'together' there are 4C2, 4!/(2!*2!)=6 combination sets

3 vowels out of 4 ones (E,A,I,O) come in 4C3, 4!/3!=4 combination sets

finally, five letter words when two (consonants) are fixed 'come together' makes 4 letter words indeed with the possibility of reversed consonants (RC-CR), hence 2 ways of 4 letter word combination sets. 4C1, 4!/1!=24 and 24*2=48

6*4*48=1152

SwatiDenre wrote:How many five letter words can be formed by picking 3 vowels and 2 consonants from the word REACTION such that the two consonants always remain together?


A. 2170
B. 2336
C. 1800
D. 1152
E. 3054


Ans - D


How to solve this prob. Please help.
Thanks in advance.

1152..

3 vowels can be selected like 4C3 = 4.
2 consonants = 4C2 = 6

5 letter word selection = 6 * 4 =24.

2 consonants must be together.ie c1c2.
3 vowels can be v1,v2,v3.
c1c2, v1,v2,v3 ie again 4 different possibilities = 4!
c1c2 can also be c2c1 ie 2 possibilities.

ie 2 * 4! orders with v1,v2,v3,v4 ,c1,c2..

5 letter selection * arranging order = 24*(2*24)

REACTION. No of consonants (C=) = 4
No. of vowels (V) = 4
select 3 vowels = 4C3 =4
select 2 consonants = 4C2 = 6
arrange (such that consonants remain together) = 4!*2! = 48
total = 4*6*48 = 24*48 = 1152
IMO D