Cool combinatorics.....
Let ( a 1 , a 2 , a 3 , … , a 1 2 ) be a permutation of ( 1 , 2 , 3 , … , 1 2 ) for which
a 1 > a 2 > a 3 > a 4 > a 5 > a 6 a n d a 6 < a 7 < a 8 < a 9 < a 1 0 < a 1 1 < a 1 2 .
How many such permutations are there?
An example of such a permutation is
( 6 , 5 , 4 , 3 , 2 , 1 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 ) .
The answer is 462.
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Since the sixth no. in the series has to be the smallest, it can only be one. Now, select any random 5 numbers from the remaining eleven (or select any 6 its the same thing) and arrange them in ascending order for the first five numbers; the rest will be the last six, which can only be arranged in one order. Hence answer is - 11C5 = 462 ...!!!