Question

1.A girl wants to buy 12 balls from toy shop. There are three type of balls.
4 red ,9 blue, 2 green
How many selections can be made if there is atleast 2 of each type of balls?

Answer 1

its actually buy 12 balls

Answer 2

just choose 2 from each category then you have to choose 6 balls from 9 balls for that just go for 9c5

Answer 3

dont worry we have commonsense :P

Answer 4

but i still don't understand that....its very confusing

Answer 5

which part?? and its an easy question in P nd C

Answer 6

my question is at the top if you get a chance please answer it

Answer 7

if only i can understand what the question means....and i don't know how to work out.

Answer 8

the only question is the understanding part.. :P

Answer 9

THIS IS FOR REVIEW PLEASE HELP I have an oral exam at 12 and need these answered help please!!! :0
How do you define the appropriate quantities to model a situation or description?
How do you choose the level of accuracy given the limitations of a situation? THIS IS FOR REVIEW

Answer 10

divu.mkr i don't get the idea .... how to do it..i need you help ... can you explain step by step

Answer 11

ok..firstly select 2 balls from each category then there will be left 2 red and 7 blue now we hav to choose 6 balls from the remaining balls and it has got a direct formula 9c6
right?? @Asha_Lvnr

Answer 12

okay, how about the 2 green balls?

Answer 13

we have already chosen dem..

Answer 14

so only 9 C 6

Answer 15

yeh.. (:

Answer 16

but the answeris 282

Answer 17

Well, sorry it took me awhile, but I'm sick today and not familiar with this subject.
However, this is how I see it:

We want to get all unique combinations of 12 balls out of those:
4 Red, 2 Green, 9 Blue.
And there must be at least 2 of each group.

That means that we can completely ignore the Green group, since there is only one possible combination of the balls in there.
Therefore, we have 10 more balls to pick out of the two remaining group (Red and Blue).

We can divide those picks into 3 unique situations, each has unique combinations (so there are no combination overlapped by two)
1) 4 Red + 6 Blue
2) 3 Red + 7 Blue
3) 2 Red + 8 Blue
That way there can't be combinations with same balls in two of those situations, because each divides the balls differently =]

Now, for each one of those situations we have to calculate the possible unique combinations that it describes.
For 1 we have:
\[
\small \bigg[\text{(Red)C4} \bigg] \cdot \bigg[\text{(Blue)C6} \bigg] \implies
\normalsize 4C4 \cdot 9C6 \implies (1) \cdot \bigg(\frac{9!}{6!} \bigg)= \large 84
\]
Because there is only one combination of Red balls and for that combination there are 84 different combinations of Blue balls.

Now to second and third:
\[
\small \bigg[\text{(Red)C3} \bigg] \cdot \bigg[\text{(Blue)C7} \bigg] \implies
\normalsize 4C3 \cdot 9C7 \implies \bigg(\frac{4!}{3! \cdot(4 - 3)!} \bigg) \cdot \bigg(\frac{9!}{7!} \bigg)= \\
\bigg(\frac{4!}{3!} \bigg) \cdot (9 \cdot 8) = (4) \cdot (9 \cdot 8) = \large 144 \\

\small \bigg[\text{(Red)C3} \bigg] \cdot \bigg[\text{(Blue)C7} \bigg] \implies
\normalsize 4C2 \cdot 9C8 \implies \bigg(\frac{4!}{2! \cdot(4 - 2)!} \bigg) \cdot \bigg(\frac{9!}{8!} \bigg)= \\
\bigg(\frac{4!}{2! \cdot 2!} \bigg) \cdot (9) = (6) \cdot (9) = \large 54 \\
\]
Now to sum all the different situations combinations together:
\[
\text{Combinations} =
\small \bigg[\text{(Red)C4} \bigg] \cdot \bigg[\text{(Blue)C6} \bigg] +
\small \bigg[\text{(Red)C3} \bigg] \cdot \bigg[\text{(Blue)C7} \bigg] + \\
\small + \bigg[\text{(Red)C2} \bigg] \cdot \bigg[\text{(Blue)C8} \bigg] =\normalsize 84 + 144 + 54 = 84 + 198 = \Large 282
\]

Honestly, there might be a shorter and more elegant way without dividing to situations, but again, i'm not familiar with the subject (yet..)