# Permutation, Combinatorics

Stuck here : there are 100 objects labeled 1, 2,...100. They are arranged in all possible ways. How many arrangements are there in which object 28 comes before object 29.

My approach : Consider object 28 & object 29 , a single object. Now we have a total of 99 objects which can be per mutated in 99! ways . But the answer is 4950*98! .

What's wrong with my approach?

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What's right with your approach? How does removing $28$ and $29$ from the pool keep track of all the different ways $28$ and $29$ can be put into the sequence (with the former coming first)? – blue Jun 25 '14 at 18:02
you are looking at the number of ways $28$ comes immediately before $29$ but the problem asks for any distance between the objects. – cirpis Jun 25 '14 at 18:04
Interestingly, the answer $4950\times 98!$ seems to have come from $\binom{100}{2}\times 98!$, so the solver did not think in terms of symmetry. – André Nicolas Jun 25 '14 at 18:09
Yup , realized my mistake . Sorry and thanks everyone :) – amrx Jun 25 '14 at 18:12
It is not really a mistake, you just interpreted the question in a different way. The English phrase "28 comes before 29" is ambiguous. – André Nicolas Jun 25 '14 at 18:24

## 4 Answers

Your answer would be right if the question asked for number of arrangements where 28 comes exactly before 29. Here, 28 can come anywhere before 29. So obviously answer would be greater. For the exact answer, note that 28 and 29 are equally likely to come before the other one. Moreover, both cannot occur together (this is obvious here, but in general it is necessary to keep this in mind). So the correct answer = 100! / 2, which can also be written as 4950 * 98!

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Without any restrictions the number of ways is $100!$. In exactly half of them $28$ will come before $29$. So the answer should be $\frac{100!}{2}.$

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Take all numbers except 28 and 29. There are $98!$ ways to arrange them.

Now place either the 28 or the 29. There are 100 places to put it. There are 99 places to put the next: $100\cdot 99 = 9900$.

Only half of those arrangements have 28 before 29, hence $\frac{100 \cdot 99}{2} = 4950$.

Multiply, getting $4950\cdot 98! = \frac{100!}{2}$.

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Your answer is correct if you mean that 28 and 29 are consecutive. However the problem does not require them to be consecutive but only that 28 is on the left of 29.

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