Permutations and Combinations. I need help.

Oct 2015
1
0
US
QUESTION: The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.

a. How many possible selections are there of four letters?
b. How many arrangements are there of four letters?


//

I know for part (a) E and S are special but I only know how to work out the answer if one letter is special.


QUESTION: Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.

a. How many different sequences of eight cards are possible?
b. How many of the sequences in part (a) will contain three picture cards, three odd numbered cards and two even numbered cards?

//

I've tried part (a) almost 10 times and my answer is still incorrect.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Part (a) is easy- since the order of the cards is irrelevant, it does not matter that some repeat. There are 9 possible choice for the first card so 8 for the second, 7 for the third, and 6 for the fourth. The number of possible four card sets is 9(8)(7)(6)= 9!/5!= 3024.

Part (b) is considerably harder since we don't want to consider different orders of the same letter, the answer depends upon how those cards are chosen.
 
May 2013
96
16
Copenhagen
Part (a) .... The number of possible four card sets is 9(8)(7)(6)= 9!/5!= 3024.

Part (b) is considerably harder since we don't want to consider different orders of the same letter, the answer depends upon how those cards are chosen.
I would say the answer is $4!\cdot 3024$ to Part (b).
 

Plato

MHF Helper
Aug 2006
22,475
8,643
QUESTION: The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.a. How many possible selections are there of four letters?
I found this to be absolutely frustrating. These are called bag problems.

I know the following solution is incorrect.
Part (a) is easy- since the order of the cards is irrelevant, it does not matter that some repeat. There are 9 possible choice for the first card so 8 for the second, 7 for the third, and 6 for the fourth. The number of possible four card sets is 9(8)(7)(6)= 9!/5!= 3024.
Change the question.
QUESTION: The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and TWO of them are selected .a. How many possible selections are there of TWO letters?
I claim the answer is $\left<< P,O >,~<P ,S >,~<P ,E >,~<O ,S >,~<O ,E >,~<S ,S >,~<S ,E >,~<E ,E >\right>$, EIGHT.

If the other logic holds the answer would be $9(8)=72$ Would it not?
Making two selections from four possibilities is $\dbinom{2+4-1}{2}=10$ but we must subtract 2 because $<P,P> \text{ or }<O,O>$ do not exist.
Clearly the OP must be much more difficult to count.