15 students in 6A are divided into 3 groups for a party , 3 students are organizers , 4 students are planners and 8 students are responsible for the food
Mandy , Helen and Sue are 6A students are responsible for the party
find the probability that thy are in the same group
thanks
+118725
(3C3 + 4C3 + 8C3)/15C3
$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$
There you go, Alan's answer and my answer look different but they work out the same because the 3! in mine cancels out.
How about that! 
+33666 This is (nbr of ways of arranging 3 people in a group of 3 + nbr of ways of arranging 3 people in a group of 4 + nbr of ways of arranging 3 people in a group of 8)/(nbr of ways of arranging 3 people in a group of 15):
$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$
.
+118725
(3C3 + 4C3 + 8C3)/15C3
$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$
There you go, Alan's answer and my answer look different but they work out the same because the 3! in mine cancels out.
How about that!
