Question

6)If you have cards numbered 1 2 3 4 5 that have to be put in 5 places

     _ _ _ _ _

and an odd numbered card must be placed on each end (cards are not replaced when used), how many arrangements are there for these cards? (Hint: use multiplication rule.)

Answer 1

solution:
Given numbered cards are 1,2,3,4,5

Let's  arrange the numbers

i) 3

ii) 3

iii) 2

iv) 1

v) 2

The No.of odd numbers from 1 to 5 = 3 - { 1,3,5 }

At Place-i) : According to the given conditions.At place-i) we need to keep an odd number from available numbers

So, No.of possible ways to keep odd numbers at place-i) = 3

At Place-v) : In this place we need to keep remaining odd numbers in 2 ways

At Place-ii) : We have 3 numbers remaining after arranging 2 odd numbers at ends. we may keep any one of 3 numbers here.So,No.of possible ways = 3

At Place-iii) : We have 2 numbers remaining.So,No.of ways to keep one number here = 2

At Place -iv) : The remaining number must be placed here,So,No.of possible ways = 1

So,Total No.of possible ways to keep odd numbers at end = 3 * 3 * 2 * 1 * 2 = 36