Mistakes give rise to Problems- 16
You can remove the brackets in a fraction (note that this is fraction, not the Legendre symbol ) , just like in the following fraction ( b a ) = b a
But if you do that in the Binomial Coefficient , like ( b a ) = b a then it's a Big Mistake !!!
But for how many ordered pairs ( a , b ) such that 0 ≤ a ≤ 2 0 and 0 ≤ b ≤ 2 0 is the above said "false" property seen to be "true" ?
Details and assumptions
∙ Factorial ( a ! ) is defined only for non-negative integers.
∙ ( b a ) = b ! × ( a − b ) ! a !
∙ ( b a ) = 0 if a < b .
∙ 0 ! = 1
This problem is a part of the set Mistakes give rise to Problems . Try other problems of the set too.
The answer is 59.
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1 solution
Note that the conclusion in the first paragraph is not entirely correct. What you have shown, is that
( ( b − 1 a − 1 ) − 1 ) × b a = 0
You only concluded that ( b − 1 a − 1 ) − 1 = 0 , but not that b a = 0 . This would explain where your third case mysteriously came from, since ( k − 1 ) = 0 .
Always remember to apply the zero product property carefully, and not ignore variables just because you can factorize them out.
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Okay sir, I will update the solution. Thanks for the awesome guidance you've always been giving ... Brilliant rocks
Here's my solution. We'll check 3 cases.
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a < b . Then equality holds if and only if a = 0 , b = 0 , which gives us 2 0 solutions.
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a = b . Then equality holds if and only if a = 0 , b = 0 . This gives us 2 0 solutions.
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a > b > 0 . Then the identity ( b a ) = b ! ( a − b ) ! a ! always holds. Our wanted equality holds if and only if ( a − 1 ) ! = ( b − 1 ) ! ( a − b ) ! holds.
(just multiply both sides by a b ! ( a − b ) ! and simplify).
This equality holds if and only if b = 1 (you can try proving this yourself or you can use a program (C++, Python, etc.)). Since a > b , we have a ≥ 2 , which gives us 1 9 solution.
Adding all the boxed numbers gives us the answer 5 9 .
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I think u missed a case (k,k-1)
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There's no variable k there. Do you mean the case when a = b + 1 (and this one is checked in the case a > b > 0 )? I don't understand what you mean, please elaborate.
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@Mathh Mathh – @mathh mathh Don't worry , he's not on the right track,
@Dipayan Bhattacharya , buddy, ( k − 1 k ) = k = k − 1 k ,
there is no equality unless k = 2 , and that case is simply ( 2 , 1 ) and it has been counted.
That's one of the uncommon mistakes, but that did give rise to a problem.
We observe that ( b a ) = b ! ( a − b ) ! a ! = b a × ( b − 1 ) ! ( a − 1 − ( b − 1 ) ) ! ( a − 1 ) ! = ( b − 1 a − 1 ) b a
Thus if you want ( b a ) = b a then you must have ( ( b − 1 a − 1 ) − 1 ) b a = 0
This gives us 2 cases, ( b − 1 a − 1 ) = 1 or a = 0
∙ This is true at all cases where a = b , (except a = b = 0 , because fraction will be not defined)
( a , b ) = ( k , k ) , 1 ≤ k ≤ 2 0 , giving 2 0 pairs.
∙ And also if b − 1 = 0 , because ( 0 n ) = n ! 0 ! n ! = 1
That is, whenever b = 1 , we have ( b a ) = ( 1 a ) = a , obviously true (again, a = 0 , we'll count the cases later )
Gives ( a , b ) = ( k , 1 ) where 1 ≤ k ≤ 2 0 , giving again 20 pairs, but ( 1 , 1 ) is repeated, so just 1 9 pairs.
∙ Other cases than this are at a = 0 . ( k 0 ) = 0 ∀ k ∈ N (number of ways of choosing k objects from 0 objects will be 0 always)
That also adds cases, ( a , b ) = ( 0 , k ) , where 1 ≤ k ≤ 2 0 , giving again 2 0 pairs, total of 2 0 + 2 0 + 1 9 = 5 9 pairs satisfying this.