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
(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!
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):
$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$
.
(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!