SATHEE: Permutations And Combinations Question 53

Permutations And Combinations Question 53

Question: How many different nine-digit numbers can be formed from the digits of the number 223355888 by rearrangement of the digits so that the odd digits occupy even places [IIT Screening 2000; Karnataka CET 2002]

Options:

A) 16

B) 36

C) 60

D) 180

Show Answer

Answer:

Correct Answer: C

Solution:

There are four even places for the four odd digits 3, 3, 5, 5. \ The required no. of ways $ =\frac{4!}{2!2!}.\frac{5!}{2!3!}=60 $ .

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Permutations And Combinations Question 53

Question: How many different nine-digit numbers can be formed from the digits of the number 223355888 by rearrangement of the digits so that the odd digits occupy even places [IIT Screening 2000; Karnataka CET 2002]

Options:

Answer:

Solution:

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