The question is as follows:

A set of 15 marbles contains 4 red and 11 green marbles. They are selected, one at a time, without replacement.
In how many ways can the last red marble be drawn on the seventh selection?

Appreciate any explanation that can be provided.

All depends if you consider you can make a difference between red marbles (for example Red1, Red2, ..., Red5) and between green marbles (G1 ... G11).

I assume here you can't (which is what usually is expected when only talking about colours).

Your problem is to fill the six first picks with 3 red marbles and 3 green marbles :

RRRGGG
RRGRGG
RRGGRG
RRGGGR

RGRRGG
RGRGRG
RGRGGR

RGGRRG
RGGRGR

RGGGRR

GRRRGG
GRRGRG
GRRGGR

GRGRRG
GRGRGR

GRGGRR

GGRRRG
GGRRGR

GGRGRR

GGGRRR

As you can see, there are 20 possibilities.

Then you must pick up a Red marble as seventh selection, which is the last red marble of the set.

You can answer such question with combinatorics : the problem is to choose 3 places among 6, for your 3 first red marbles.

"3 among 6" is a combinatoric number, equal to :

6! / (3! * (6-3)!) = 6*5*4 / (3*2*1) = 5*4 = 20

This post was edited by feanur on Jan 28 2017 12:26am

Wow what an excellent explanation Thank you!