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The order of a leaf node in a $\text{B tree}$ is the maximum number of (value, data record pointer) pairs it can hold. Given that the block size is $1 \mathrm{~K}$ bytes, data record pointer is $7$ bytes long, the value field is $9$ bytes long and a block pointer is $6$ bytes long, what is the order of the leaf node?
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Ohhk @Neha24  Thanku 

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In the question, the definition of order has been changed. Let $\mathrm{P}$ be the order then we have : $\mathrm{P}(7+9)+(\mathrm{P}+1) 6\leq1024$

$\mathrm{P}=46$.
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Why block pointer is considred here?..leaf node of btree contains only key and record pointer.

@GO Classes

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In a B-tree of order $n$, all nodes including leaf contains $n$ block pointers and $(n-1)$ key and record pointers
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but the general defination is p*Bp + (p-1)(k+Rp) <= size  ???
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ans of that question is 47 not 46 because we consider the floor of N (order of tree) so its 47.2727, so its 47... @Anyone please clear if something is worng
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@Aman_777  I also did same mistake:( In given qus the definition of order has been changed.

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@Deepak Poonia sir, does root node of a b tree contains block pointers
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@Omkar Rasal  replace p with p+1 because here question demands the maximum between  record pointer and key pointer size ..which is clearly key value .

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@Rahulji  yes all nodes from root to leaf contains Bp.

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Answer:

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