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The number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time such that digit 1 appearing somewhere to the left of 2, 3 appearing to the left of 4, and 5 somehere to the left of 6 is(eg 815723945 would be one such permutation)

The number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time such that digit 1 appearing somewhere to the left of 2, 3 appearing to the left of 4, and 5 somehere to the left of 6 is(eg 815723945 would be one such permutation)

Grade:10

1 Answers

Arun
25763 Points
3 years ago
Dear student
 
There are $9!$ (the factorial of 9) total permutations of the elements of that set. 1 is to the left of 2 in exactly half of these. 3 is also to the left of 4 in exactly half of the permutations, and 5 is to the left of 6 in exactly half of the permutations. These three events are totally independent of each other, so the number we want to calculate is $\frac12\cdot\frac12\cdot\frac12\cdot9! = \frac18\cdot9\cdot8\cdot7! = 9\cdot7!$ 
 
Regards
Arun(askIITians forum expert)

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