No solution?
Find the integer n = 2 0 1 5 such that
( 1 4 n ) − ( 1 4 2 0 1 5 ) = 0 .
The answer is -2002.
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1 solution
steven....nCr is defined only for n belonging to natural no.
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Yes, that is what made me to dump the idea of n being a negative number.
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I think I should list this problem under Algebra instead, because the solution is algebraic rather than related to actual combinations or counting.
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@Steven Yuan – @steven Yaun:but the operations you applied are completely related with combinations.there is no way you can apply it to rather normal algebra
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@Mudit Bansal – The binomial coefficients have a natural extension to negative and even fractional coefficients.
They do have applications in combinatorics. For example, they can be obtained by "extending" the Pascal's triangle up above the first row.
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@Calvin Lin – you can apply -ve and fractional coefficients to n in (a+bx)^n but not in nCr.nCr is by defualt defined for n belonging to natural no.
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@Mudit Bansal – Yes and so I think because both the logic are correct in their sense, both are correct.
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@Kartik Sharma – Well, we can define ( k n ) so that n can be negative or fractional. Since
( k n ) = k ! ( n − k ) ! n ! ,
we can divide the numerator and denominator by ( n − k ) ! to get
( k n ) = k ! ( n ) ( n − 1 ) ⋯ ( n − k + 1 ) .
This definition allows us to calculate ( k n ) when n is negative or fractional; the only requirement is that k must be a nonnegative integer.
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@Steven Yuan – interesting! can you further explain how to extend the pascals triangle in the upward direction then?
WA'ed this lol
Rewrite the equation as
( 1 4 n ) = ( 1 4 2 0 1 5 ) .
Since n = 2 0 1 5 , n has to be negative. (Yes, there is such a thing!) To find n , we use the fact that, for positive integers a and b ,
( b − a ) = ( − 1 ) b ( b a + b − 1 ) .
In this case, we have b = 1 4 and a = 2 0 0 2 , so n = − 2 0 0 2 .