Question

find the number of permutation of the letters of the word REMAINS such that the vowels always occur in odd places​

Answer 1

Step-by-step explanation:

Odd places are total 4 i.e 1,3,5,7 and vowels are 3 so 4p3 =6 ways

Even places for four consonants 4 consonants 4p4 =24

Total 6×24=144 ways

Answer 2

Step-by-step explanation:

• There are 7 letters in the word REMAINS, out of which there are 3 vowels and 4 consonants.
• Now in 4 different places, the 3 vowels can occur. So it will be
•                                 4 P3 ways
•                                 4! / (4 – 3)!
•                                   4!
•                                   4 x 3 x 2 x 1
•                                       24 ways.
• Also 4 consonants can take 4 places. So it will be
•                                                             4P 4 ways
•                                                                           4!
•                                                                           24 ways
• So the number of permutations possible will be 24 x 24 = 576 ways.

Reference link will be

https://brainly.in/question/37836381?m