# [R] Permutations

R. Michael Weylandt michael.weylandt at gmail.com
Sat Nov 19 02:25:02 CET 2011

```What build of R can't calculate factorial(150)? I thought the max
double of a 32 bit build would be on the order of 10^308 ~ 2^1024 (the
material below seems to agree with me on this)

But yes, agreed with Ted: it's very helpful to think a bit about what
you are calculating when doing these sorts of large number
calculations before asking for the accurate calculation of huge
numbers.

Michael

On Fri, Nov 18, 2011 at 4:31 PM, Ted Harding <ted.harding at wlandres.net> wrote:
> On 18-Nov-11 16:03:44, Gyanendra Pokharel wrote:
>> Hi all,
>> why factorial(150) shows the error out of range in 'gammafn'?
>> I have to calculate the number of subset formed by 150 samples
>> taking 10 at a time. How is this possible?
>> best
>
> Because factorial(150) is nearly 10^263, which is far greater
> than any number that R is capable of storing.
>
> I would, perhaps, suggest you work with lgamma(), the logarithm
> (to base e) of the Gamma function.
>
> Thus factorial(150) = gamma(151), and
>
>  lgamma(151)
>  # [1] 605.0201
>
> so factorial(150) is close to e^605; to get it as a power of 10:
>
>  lgamma(151)*log10(exp(1))
>  # [1] 262.7569
>
>
> Hence the "nearly 10^263" above.
>
> If your question means "the number of ways of choosing 10 out
> of 150, then, using lgamma(), its log to base e is:
>
>  lgamma(151) - lgamma(11) - lgamma(141)
>  # [1] 34.6954
>
> and its log to base 10 is:
>
>  (lgamma(151) - lgamma(11) - lgamma(141))*log10(exp(1))
>  # [1] 15.06802
>
> so the result is about 10^15 which *is* within R's range.
>
> Hence:
>
>  10^( (lgamma(151) - lgamma(11) - lgamma(141))*log10(exp(1)) )
>  # 1.169554e+15
>
> You can see this (approximately) in an all-digits form by
> using print():
>
>  X <- 10^( (lgamma(151) - lgamma(11) - lgamma(141))*log10(exp(1)) )
>  print(X,18)
>  # [1] 1169554298222353
>
> However, in many cases (such as your example) it will be feasible
> to exploit the reduction in numbers of factors if you cancel out
> the factors in the larger denominator factorial from the numerator
> factorial, so that the number you want is
>
>  150*149*148*147*146*145*144*143*142*141/(10*9*8*7*6*5*4*3*2*1)
>
> = (150/10)*(149/9)*(148/8)* ... *(142/2)*(141/1)
>
> for which you can define a function nCr(n,r):
>
>  nCr <- function(n,r){
>    num <- (n:(n-r+1))
>    den <- (r:1)
>    prod(num/den)
>  }
>
> and then:
>
>  print(nCr(150,10),18)
>  # [1] 1169554298222310
>
> which is not quite the same as the result 1169554298222353
> obtained above using lgamma(). This is because the previous
> result is affected by rounding errors occuring in the
> logarithms. The result obtained using the function nCr() is
> exact. However, it can fail in its turn if you want one of
> the larger possible results, such as nCr(1000,500):
>
>  nCr(1500,750)
>  # [1] Inf
>
> since the result is too large for R to store (the largest on
> my machine is about 2*10^308, see below)), and the products
> in the function definition accumulate until this number is
> exceeded.
>
> You can find out the limits of R storage on your machine
> by using the command
>
>  .Machine
>
> which gives a long list of the characteristics of the machine
> you are using. In particular, you will see:
>
>  \$double.xmax
>  [1] 1.797693e+308
>
> is the largest number that R will store; and:
>
>  \$double.eps
>  [1] 2.220446e-16
>
> which is the smallest positive number; and:
>
>  \$double.neg.eps
>  [1] 1.110223e-16
>
> which is the smallest negative number (note that the latter
> is half the former).
>
> They can be individually accessed as:
>
>  .Machine\$double.xmax
>  # [1] 1.797693e+308
>
>  .Machine\$double.eps
>  # [1] 2.220446e-16
>
>  .Machine\$double.neg.eps
>  # [1] 1.110223e-16
>
> Hoping this helps!
> Ted.
>
> --------------------------------------------------------------------
> E-Mail: (Ted Harding) <ted.harding at wlandres.net>
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> Date: 18-Nov-11                                       Time: 21:31:04
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