Let `nbbHr` be a symbolic notation for the number of possible combinations
in
selecting `r` samples out of a pool of `n` different things with replacing
the
particular sample by the same thing in the pool every time it is taken
out.
The values can be obtained visually by using a rectangle method.
(Ignore
the parenthesis of a matrix notation in the rectangle below.)`stackrel()(->)
stackrel()(->) stackrel()(->) stackrel()(->) stackrel()(->) Column: r=1, 2, 3,
4, cdots` `darr Row: n=1,2,3,4,cdots
((1,1,1,1,cdots),(2,3,4,5,cdots),(3,6,10,15,cdots),(4,10,20,35,cdots),(.,.,.,.,c
dots))` where the first row is all `1` because there is only one way to
select samples
from a pool of one for any `r`, the first culumn is
`1,2,3,4,cdots` because there are
`n` ways to select one sample from a pool of
`n`, and the other `nbbHr` values can
be constructed by adding its left number
and its above number in the rectangle.
For example, `2bbH2 = 2bbH1 + 1bbH2 =
2+1 = 3`,
`2bbH3 = 2bbH2 + 1bbH3 = 3+1 = 4`,
`3bbH2 = 3bbH1 + 2bbH2 = 3+3
=6`,
`3bbH3 = 3bbH2 + 2bbH3 = 6+4 = 10`, and so on. In fact, the above
rectangle is a part of Pascals triangle rearranged into
a rectangle shape. If
we allow `r=0` by defining `nbbH0=1` for all positive
integers `n`, then the
rectangle may look more familiar. `((1, 1, 1,1,1,cdots),(1, 2,
3,4,5,cdots),(1, 3, 6,10,15,cdots),(1, 4,10,20,35,cdots),(1,
5,15,35,70,cdots),(.,.,.,.,.,cdots))` 1) Confirm the above values using
explicit formulas
(i.e., `nbbHr = (n+r-1)bbCr =
((n+r-1)bb!)/(rbb!(n-1)bb!)`). 2) Prove the following equation, which the
above rectangle method bases upon,
without using explicit formulas: `nbbHr =
nbbH(r-1) + (n-1)bbHr`
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