Just give it a think

Probability Level 5

There are $2015$ seats numbered $1,2,3,\cdots,2015$ .

There are also $2015$ persons who are holding tickets $1,2,3,\cdots,2015$ .

They have seats at random.

Then the probability that exactly $3$ persons take seats having same numbers as that on their tickets is $P$ .

Find $\lfloor10^{6}P\rfloor$ .

The answer is 61313.

2 solutions

Mvs Saketh
Feb 23, 2015

Select any 3 persons in 2015C3 ways,

Now derange the rest 2012 , but it is not possible to compute such a large derangement, But fortunately, the derangement formula is simply the partial expansion of n!e^(-1) or n!/e and it converges very rapidly with Derangement of numbers,

infact D(4) = 9 and , and 24/e = 8.829 which is pretty close,

For 2012, this is almost exactly close to 2012!/e

using it we claim

P = (2015C3)(2012!/e) / (2015!) = 1/6e

multiplying by 10^6 gives the answer

Subrata Dutta
Feb 22, 2015

first we select 3 persons from 2015 persons in 2015C3 ways . Each of these 3 persons can sit only in 1 position . And the rest 2012 persons will sit in the rest 2012 seats so that none will seat in write position in (1/2! -1/3! +1/4! -1/5! +1/6! -1/7! +.......+1/2012!) *2012!=A ways. So the required probability

=(2015C3*A)/2015! =0.06131324..

I just wanted to know how did you evaluated the value of A without calculator

Vighnesh Raut - 6 years, 3 months ago

Just give it a think

The answer is 61313.

2 solutions

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