# Combinatorics Questions (1 Viewer)

#### SB257426

##### Very Important User
I just started learning this topic so please don't laugh at me if you find this question easy but I can't do any of these .....

#### Average Boreduser

##### Rising Renewal
I just started learning this topic so please don't laugh at me if you find this question easy but I can't do any of these .....

View attachment 38776
Trust me bro, no one is gonna judge. Perms and combs are cancerous.

#### synthesisFR

##### Well-Known Member
Nah I would fail this
.. idk what’s in a deck of cards don’t judge

#### cossine

##### Well-Known Member
I just started learning this topic so please don't laugh at me if you find this question easy but I can't do any of these .....

View attachment 38776
Most questions seem to be employing rule of product axiom. So as a first start apply the axiom.

#### Average Boreduser

##### Rising Renewal
Most questions seem to be employing rule of product axiom. So as a first start apply the axiom.
Whats that?

#### carrotsss

##### New Member
Once you've read through a couple and know the general method, try to give each individual question a go before reading its answer. This style of questions is quite important to nail down -- obviously the card terminology isn't that important but you do need to be able to extract information from questions well for this topic.

a) 4, there's only one possibility for each suit
b) You can start with A,2,3,4,5,6,7,8,9,10, so 10 possibilities to begin with. Because the suit doesn't matter, each individual combination has 4^4 possible combination of suits. Therefore, 2560
c) 4 suits, 10 possibilities for each -> 40
d) For each suit we have 13 cards, and we need to pick 5 cards, and order doesn't matter. Therefore, 4 (suits) * 13C4 (cards within suit) = 2860
e) First we pick which card theres going to be 4 of, so 13 different possibilities. Then we have 48 cards left to choose from, so 48*13=624
f) First, we pick the one which have 3-of-a-kind, so 13 possibilities, and then we need to figure out which suit will be excluded, so multiplied by 4. Similarly, we need to pick the pair out of the remaining 12 possibilities, and then pick which 2 suits are excluded, so 4C2. Hence, 13*4*12*4C2=3744
g) We pick the 2 pairs (so 13C2) and then which cards will be excluded in each pair (4C2*4C2). Then, we have 44 cards left to choose from for the rest (can't use something from one of the numbers we've already got), so 13C2*4C2*4C2=2808
h) Just pick what the pair will be (so 13*4C2 just like the previous questions) and then we have 48 cards remaining, so 48C3 for the rest of the cards, so 13*4C2*48C3=1349088
i) First, we pick the number (13 choices), then which one will be excluded (4 choices), then 48 cards left and we need to pick 3 and order doesn't matter so 13*4*48C3=899392

Apologies if I got anything wrong, I rushed this and I haven't done one of these questions in a while.

#### carrotsss

##### New Member
Nah I would fail this
.. idk what’s in a deck of cards don’t judge
4 suits; hearts, clubs, diamonds, spades
Each suit has an A,2,3,4,5,6,7,8,9,10,J,Q,K

#### howcanibesmarter

##### Well-Known Member
I feel like with these types of questions, its so hard to get started, but once you see the solutions its makes so much sense that you question how you didn't even get it.