Permutations & Combinations
example problems
Example problems
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?
Answer (highlight to read):
The number of arrangements of 9 unique cards would be 9!.
Because of the repeated cards the actual number of arrangements is 9! / ( 5!·2! ).
To find the number of possible 4 card selections this is further reduced by dividing by the number of ways of picking a set of 4 cards from 9.
Hence the number of selections is equal to 9! / ( 5!·2!·^{9}C_{4} ).
There are thus 12 selections possible.
Since S can be repeated to make up all four cards, I'll list the possible selections alphabetically, and according to the number of Ss they include.
0S: EEOP
1S: EEOS EEPS EOPS
2S: EESS EOSS EPSS OPSS
3S: ESSS OSSS PSSS
4S: SSSS
b) How many arrangements are there of four letters?
Answer (highlight to read):
For each selection there are a certain number of possible arrangements which varies according to how many repeated letters there are in the particular selection.
0S: EEOP ⇒ 4! / 2! = 12 (Alternatively, the double letters can be in each of six arrangements in relation to the other two letters, which can themselves be arranged two different ways, doubling the number of arrangements.)
1S: EEOS ⇒ 4! / 2! = 12
EEPS ⇒ 4! / 2! = 12
EOPS ⇒ 4! = 24
2S: EESS ⇒ 4! / ( 2!·2! ) = 6
EOSS ⇒ 4! / 2! = 12
EPSS ⇒ 4! / 2! = 12
OPSS ⇒ 4! / 2! = 12
3S: ESSS ⇒ 4! / 3! = 4 (Alternatively, the single letter can be in each of four positions,)
OSSS ⇒ 4! / 3! = 4
PSSS ⇒ 4! / 3! = 4
4S: SSSS ⇒ 1
There are thus 12 * 6 + 24 + 6 + 4 * 3 + 1 = 72 + 30 + 13 = 115 arrangements of the cards.
