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Permutations and Combinations question

2023 · 24 Jan · Shift 1 · Q44
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Permutations and Combinations question

2023 · 24 Jan · Shift 1 · Q44

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 60

  1. Digits available

The number is 123412341123412341123412341.

So the multiset of digits is: 1,1,1,2,2,3,3,4,41,1,1,2,2,3,3,4,41,1,1,2,2,3,3,4,4

Thus:

  • Odd digits: 1,1,1,3,31,1,1,3,31,1,1,3,3
  • Even digits: 2,2,4,42,2,4,42,2,4,4

We have to form 9-digit numbers using all these digits such that even digits occupy only even places.

  1. Available positions

In a 9-digit number, the positions are: 1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9

Even places are: 2,4,6,82,4,6,82,4,6,8

Odd places are: 1,3,5,7,91,3,5,7,91,3,5,7,9

Since all 4 even digits (2,2,4,42,2,4,42,2,4,4) must occupy only even places, they must be placed in positions 2,4,6,82,4,6,82,4,6,8.

Similarly, the 5 odd digits (1,1,1,3,31,1,1,3,31,1,1,3,3) must occupy positions 1,3,5,7,91,3,5,7,91,3,5,7,9.

  1. Arrange even digits in even places

We need to arrange 2,2,4,42,2,4,42,2,4,4 in 4 distinct even positions.

Number of distinct arrangements: 4!2!2!=6\frac{4!}{2!2!} = 62!2!4!​=6

  1. Arrange odd digits in odd places

We need to arrange 1,1,1,3,31,1,1,3,31,1,1,3,3 in 5 distinct odd positions.

Number of distinct arrangements: 5!3!2!=10\frac{5!}{3!2!} = 103!2!5!​=10

  1. Total number of valid 9-digit numbers

By multiplication principle, 6×10=606 \times 10 = 606×10=60

  1. Comparison with stored answer

Our derived answer is: 606060

The stored correct answer is also 606060.

So they agree.

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