Combinatorics practice problems

Q1: At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

Answer: 18

Solution:
Each of the 12 men danced with 3 women each.
Total number of dances performed = 12*3 = 36.

Each of the X women danced with 2 men each.
36 = X*2 => X = 18.

Q2. A sandwich shop offers its sandwiches with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three bread patties, and any collection of condiments. How many different kinds of sandwiches can be ordered?

Answer: 768
Solution:
1. Customer can choose 0,1,2,3,4,5,6,7,8 condiments.
2. Each choice again corresponds to 1,2,3 patties.
For 1. 8C0 + 8C1 + ... 8C8 = 256
Why?
Using Binomial theorem
(x + y)^n = nC0.x^n.y^0 ..... nCn.x^0.y^n
Put x = y = 1 => 2^n = nC0 + ... nCn
So for 8 it's 2^8 = 256.
256*3 = 768.

Another way to see it is this:
Each condiment can be chosen or not chosen. So 2 options per condiment.
2*2... 8 times = 2^8 = 256.

Q3: A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only 1/3 of the marbles in the bag are blue. Then yellow marbles are added to the bag until only 1/5 of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

Answer: 1/3
Solution:
Initially B > R.
Then B = x, R = 2x.
Then B = x, R = 2x, Y = 2x.
Then B = 2x, R = 2x, Y= 2x.
So 1/3.

Q4: India changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?
Answer: (26^2)/10
Solution:
Earlier: 26*10^4
Now: 26^3*10^3
Ratio of Now/Earlier is the answer.

Q5: Sameer wants to buy four Pastries from an ample supply of three types of Pastries : glazed, chocolate, and powdered. How many different selections are possible?
Answer: 15
Solution:
Straightforward application of stars and bars method.
n = 4, k = 3
(n+k-1)C(k-1) = 6C2 = 15





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