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Q. In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?

VITEEEVITEEE 2008

Solution:

We know that the number of ways of dividing (m+n+p) things into three groups containing m, n and p things respectively
$\quad$$\quad$$\quad$$=\frac{\left(m+n+p\right)!}{m!n!p!}$
Further if any two groups out of the three have same number of things then number of ways
$\quad$$\quad$$\quad$$=\frac{\left(m+n+p\right)!}{m!n!p!\times2}$
Hence number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three
$=\frac{10!}{4!3!3!\times2}=\frac{10\times9\times 8\times 7\times 6\times 5}{3\times 2\times 3\times 2\times 2}=2100$