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By pigeonhole principle, show that if any five numbers from 1 to 8 are chosen, then two of them will add upto 9.
Answer: Let make four groups of two numbers from 1 to 8 like that sum the numbers in a group is 9. The groups are as following: (1, 8), (2, 7), (3, 6) and (4, 5).
Let us refer these four groups like pigeonholes (m). So m = 4. Take the five numbers to be choosen arbitrarily as pigeons that is n = 5. Take a pigeon and put in the pigeonhole according to its value. After placing 4 pigeons, the 5th has to go in one of the pigeonhole. That is by pigeonhole principle has at least one group that will contain [(5-1)/4]+ 1 numbers. So two of the numbers, out of the five selected, will add up to 9.
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If roots of (x-p)(x-q) = c are a and b what will be the roots of (x-a)(x-b) = -c please explain. Solution) (x-p)(x-q)=c x2-(p+q)x-c=0 hence, a+b=p+q and a.b=pq-c
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