This is a regular Permutation Combination problem
Out of 5 oil paintings, he chooses 3. that is 5C3 = 5!/2!3! = 10
Out of 6 water color, he choose 4 = 6!/4!2! = 15

Since he has to choose both oil and water color, the answer is 15 * 10 = 150

Relaxin,

Choosing k things from n things available (usually involves either permutations or combinations)

Here the order does not matter. Why? Lets say there are 5 paintings
A,B,C,D,E and I have to choose 3. If I choose A,B,D its the same as me choosing B,A,D or D,A,B or D,B,A or A,D,B or B,D,A.This applies to any set of 3 choosen

So its C(5,3) = n C r = 5!/3!*2! = 10

Silimarly for the 6 water colors its 6C4 = 15

Now I have 10 different combinations of paint and 15 different combinations of water colors i.e. choosing the paint can happen in 10 different ways(lets say m) and each of these m ways can be followed by choosing the watercolor in 15 different ways (lets say n) then by the fundamental counting principle the 2 events can happen in m*n ways (i.e 10*15 = 150)

Hope this helps!

Another alternate expln:

This is a combination problem not a permutation one, as the order of paintings here is not important.
i.e. in the exhibition, there is no mention of ordering of the paintings, or there's no restriction/ condition either.
So, it's a combination problem.

From now, on it's straighfwd:
the valid combinations are: 5C3 * 6C4 = (5!/(3!2!)) * (6!/(4!2!))
= 10 * 15 = 150

Permutation: order matters n!/(n-k)!
Combination: order does not matter n!(n-k)!k!

Combination problem 5!/(5-3)!3! = 5!/2!*3! = 3*4*5/3 *2 = 60/6 = 10
6!/(6-4)!4! = 6!/2!*4! = 3*4*5*6/4*3*2 = 360/24 = 15

In this case, the 2 cancels below leaving 3*4*5*6 in the numerator to be divided by 4 * 3 * 2 in the denominator


or alternatively
6!/2!4! = 5 * 6 /2 *1 = 30/2=15

In this case, the 4 cancels and that leaves 5 * 6 in the numerator to be divided by 2 factorial or 2 = 15



10 * 15 = 150 possible combinations

Permutation: order matters n!/(n-k)!
Combination: order does not matter n!(n-k)!k!

Combination problem 5!/(5-3)!3! = 5!/2!*3! = 3*4*5/3 *2 = 60/6 = 10
6!/(6-4)!4! = 6!/2!*4! = 3*4*5*6/4*3*2 = 360/24 = 15

In this case, the 2 cancels below leaving 3*4*5*6 in the numerator to be divided by 4 * 3 * 2 in the denominator


or alternatively
6!/2!4! = 5 * 6 /2 *1 = 30/2=15

In this case, the 4 cancels and that leaves 5 * 6 in the numerator to be divided by 2 factorial or 2 = 15



10 * 15 = 150 possible combinations

thank you all so much, you know permutations and combinations are pretty much the only thing ive had trouble with, skeee