SATHEE: Permutations And Combinations Question 374

Permutations And Combinations Question 374

Question: How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Options:

A) 18

B) 28

C) 6

D) 27

Show Answer

Answer:

Correct Answer: A

Solution:

[a] There are 4 odd digits (1, 1, 3, 3) and 4 odd place (first, third, fifth and seventh). At these places the odd digits can be arranged in $ \frac{4!}{2!2!} $ =6 ways Then at the remaining 3 places, the remaining three digits (2, 2, 4) can be arranged in $ \frac{3!}{2!}=3 $ ways
$ \therefore $ The required of number of numbers $ =6\times 3=18 $

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Permutations And Combinations Question 374

Question: How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?

Options:

Answer:

Solution:

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