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question

Agnishom has 10 pockets and 44 coins . He wants to put his coins into his pockets so distributed that each pocket contains a different number of coins

  1. Can he do so?

  2. Generalize the problem,considering p pockets and n coins.

The problem is most interesting when

\( n = \frac{(p + 1)(p - 2)}{2}\)

Why?

Note by Megh Choksi
2 years, 2 months ago

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  1. He can not do so. Because He can't keep \(2\) or more pockets empty (as it is given that each pocket has different number of coins) . And at the least, if we keep one pocket empty, then also minimum numbers of coins required to put in \(9\) pockets is \(\displaystyle \sum_{i=1}^{9}=45\).

  2. Minimum number of coins required to put into p pockets (according to the restrictions applied) \(=\dfrac{p(p-1)}{2}\)

Sandeep Bhardwaj · 2 years, 2 months ago

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@Sandeep Bhardwaj yes Sir its right , can you answer last part Megh Choksi · 2 years, 2 months ago

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@Megh Choksi Same is the case with the last part. Minimum no. of coins required \(=\dfrac{p(p-1)}{2}\). And as you mentioned there , if no. of coins is \(\dfrac{(p+1)(p-2)}{2}\) which is one less than the no. of coins required. I want to say that \(\dfrac{p(p-1)}{2}-\dfrac{(p+1)(p-2)}{2}=1\). So, in this case too, it is not possible to put the coins into the pockets with the restrictions applied. @megh choksi Sandeep Bhardwaj · 2 years, 2 months ago

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@Sandeep Bhardwaj Yes right , thank you for replying Megh Choksi · 2 years, 2 months ago

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1+2+3+4+5+6+7+8+9=45. He will have one empty pocket! Venture Hi · 2 years, 2 months ago

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@Venture Hi Sorry @Venture HI , I was waiting for your response to the second part . nicely done ,voted up thank you Megh Choksi · 2 years, 2 months ago

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