" Find the number of ways of giving 10 identical gift boxes to 6 people : A, B, C, D, E, F in such a way that total number of boxes given to A and B together does not exceed 4. "
I tried it in this way :
[$x^{10}$] $(1+x^{1}+...+x^{4})^{2}*( 1+x^{2}+....)^{4}$
But I am not sure if its right or not , can anyone help me with the condition on A and B i.e say $x_1+x_2\leq 4$
What you've to find is $x_1+x_2+x_3+x_4+x_5+x_6=10$ (I've replaced A-F by 1-6). Further, $x_i\geq 0$ and $x_1+x_2\leq 4$.
Required answer= $$\sum_{i=0}^{4}(Coeff.\ of\ x^i\ in \ (x^0+x^1+\cdots x^4)\cdot (x^0+x^1+\cdots x^4))\cdot(Coeff.\ of \ x^{10-i}\ in \ (x^0+x^1+\cdots x^{10})^4)$$ Can you solve it now?
In open form, it can also be written as: $$(Coeff.\ of\ x^0\ in \ (x^0+x^1+\cdots x^4)^2)\cdot(Coeff.\ of \ x^{10}\ in \ (x^0+x^1+\cdots x^{10})^4)$$$$+(Coeff.\ of\ x^1\ in \ (x^0+x^1+\cdots x^4)^2)\cdot(Coeff.\ of \ x^{9}\ in \ (x^0+x^1+\cdots x^{10})^4)$$$$+(Coeff.\ of\ x^2\ in \ (x^0+x^1+\cdots x^4)^2)\cdot(Coeff.\ of \ x^{8}\ in \ (x^0+x^1+\cdots x^{10})^4)$$$$+(Coeff.\ of\ x^3\ in \ (x^0+x^1+\cdots x^4)^2)\cdot(Coeff.\ of \ x^{7}\ in \ (x^0+x^1+\cdots x^{10})^4)$$$$+(Coeff.\ of\ x^4\ in \ (x^0+x^1+\cdots x^4)^2)\cdot(Coeff.\ of \ x^{6}\ in \ (x^0+x^1+\cdots x^{10})^4)$$