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Permutations and Combinations question

2023 · 10 Apr · Shift 1 · Q43
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Permutations and Combinations question

2023 · 10 Apr · Shift 1 · Q43

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4898

  1. Total permutations of 1,2,3,4,5,6,71,2,3,4,5,6,71,2,3,4,5,6,7

Since all 7 digits are distinct, the total number of permutations is 7!=5040.7! = 5040.7!=5040.

  1. Count permutations containing the string 153153153

If the string 153153153 appears consecutively, treat it as one block.

Then the objects are: [153],2,4,6,7[153], 2,4,6,7[153],2,4,6,7 So we have 555 objects in total.

Hence the number of permutations containing 153153153 is 5!=120.5! = 120.5!=120.

  1. Count permutations containing the string 246724672467

If the string 246724672467 appears consecutively, treat it as one block.

Then the objects are: 1,3,5,[2467]1,3,5,[2467]1,3,5,[2467] So we have 444 objects in total.

Hence the number of permutations containing 246724672467 is 4!=24.4! = 24.4!=24.

  1. Count permutations containing both 153153153 and 246724672467

Now treat both as blocks: [153],[2467][153], [2467][153],[2467] There are no remaining digits.

So the number of permutations containing both strings is just the number of ways to arrange these two blocks: 2!=2.2! = 2.2!=2.

These are: 1532467,2467153.1532467, \quad 2467153.1532467,2467153.

  1. Apply inclusion-exclusion

Number of permutations containing at least one of the two strings is 120+24−2=142.120 + 24 - 2 = 142.120+24−2=142.

Therefore, the number of permutations containing neither string is 7!−142=5040−142=4898.7! - 142 = 5040 - 142 = 4898.7!−142=5040−142=4898.

  1. Final answer

4898\boxed{4898}4898​

  1. Comparison with stored answer

Stored correct answer = 489848984898.

Our derived answer matches the stored answer.

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