No CrossRef data available.
Published online by Cambridge University Press: 20 January 2009
The number of groups of n which may be selected from 2n is 2n(2n − l)…(n + l)/n! But make the 2n into two groups of n, and select r out of the first and n − r out of the second. This gives [n(n−l)…(n−r+l)/r!] + [n(n−l)…(r + l)/(n−r)!] ways of thus making a group of n. Hence
* The use of this identity was suggested to me by Professor Tait.
* Compare a paper by MrLeudesdorf, , in the Proceedings of the Lond. Math. Soc. for 1889, p. 199–a paper which I did not see till after the above was written.Google Scholar