More fun in 2015, Part 31

Probability Level 4

Consider a sequence { a i } \{ a_i \} of distinct arbitrary natural numbers of length 2013.

Find the number of possible sequences of natural numbers of length 2013 2013 such that 1 ≤ a 1 < a 2 < a 3 < ⋯ < a 2012 < a 2013 ≤ 2015 1\leq a_1 < a_2 < a_3 < \cdots < a_{2012}<a_{2013} \leq 2015

Bonus: Generalize this.


The answer is 2029105.

Nihar Mahajan by Nihar Mahajan , 20, India

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1 solution

Kay Xspre
Nov 20, 2015

From 1 to 2015 inclusive, there are ( 2015 2013 ) = 2015 ! 2 ! 2013 ! = 2 , 029 , 105 \binom{2015}{2013} = \frac{2015!}{2!2013!} = 2,029,105 ways to pick 2013 numbers to form a sequence. Since the number can be in any order without restriction (consecutive, etc...), we came to that conclusion.

Generalized form is from x x to y y inclusive, selecting z z number to form the sequence with properties that a n < a n + 1 a_n < a_{n+1} (except a 1 a_1 and the last term) will contain ( y − x + 1 z ) \binom{y-x+1}{z} possible ways

1 Helpful 0 Interesting 0 Brilliant 0 Confused

If the sequence also included 0 (i.e a i ∈ W a_i \in W ) and inequality reversed then a good problem can be made.

Gautam Sharma - 5 years, 6 months ago

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