引理：(a1+a2+...+ak)^n的展开式中a1^n1*a2^n2*...*ak^nk项（n1+n2+...+nk=n）的系数是n!/(n1!*n2!*...*nk!).
证明：对k用归纳法：
k=1时显然.
假设对k成立,要证对k+1成立.注意到
(a1+a2+...+a{k-1}+ak+a{k+1})^n=(a1+a2+...+a{k-1}+(ak+a{k+1}))^n.
把ak+a{k+1}看成一项,则由归纳假设知展开式中a1^n1*a2^n2*...*a{k-1}^n{k-1}*(ak+a{k+1})^nk项（n1+n2+...+nk=n）的系数是n!/(n1!*n2!*...*n{k-1}!*nk!).
再由二项式定理,(ak+a{k+1})^nk展开式中ak^mk*a{k+1}^m{k+1}项（mk+m{k+1}=nk）的系数是(nk取mk)=nk!/(mk!*m{k+1}!).
所以(a1+a2+...+a{k-1}+ak+a{k+1})^n的展开式中a1^n1*a2^n2*...*a{k-1}^n{k-1}*ak^mk*a{k+1}^m{k+1}项（n1+n2+...+n{k-1}+mk+m{k+1}=n）的系数是
[n!/(n1!*n2!*...*n{k-1}!*nk!)]*[nk!/(mk!*m{k+1}!)]
=n!/(n1!*n2!*...*n{k-1}!*mk!*m{k+1}!).
证毕.
所以所求系数=9!/(2!*3!*4!)=1260