English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

a)How many possible selections are there of 4 letters?
b) How many arrangements are there of 4 letters?
( Permutations & Combinations)

2007-01-21 06:27:24 · 3 answers · asked by jay 1 in Science & Mathematics Mathematics

3 answers

To answer a), after futzing with trying to do it using combinations for most of an hour, I've decided just to list them all out.

There are only 4 different letters, of which P and O can only be chosen once, E can be chosen once or twice, and S can be chosen up to 4 times. (The fact that there are 5 S's doesn't matter, because we can only use at most 4 of them.)

For each selection of 4 letters, the number of arrangements is 4! if there are no repeated letters. If there are repetitions, we have to divide by the factorial(s) of the number of repetitions in order to eliminate identical arrangements.

Using both P and O:
POEE: 4!/2! = 12 arrangements
POES: 4! = 24 arrangements
POSS: 4!/2! = 12 arrangements

Using P:
PEES: 4!/2! = 12 arrangements
PESS: 4!/2! = 12 arrangements
PSSS: 4!/3! = 4 arrangements

Using O:
OEES: 4!/2! = 12 arrangements
OESS: 4!/2! = 12 arrangements
OSSS: 4!/3! = 4 arrangements

The rest:
EESS: 4!/(2!2!) = 6 arrangements
ESSS: 4!/3! = 4 arrangements
SSSS: 1 arrangement

So, I believe there are 12 possible distinct selections, and (12 + 24 + 12 + 2(12 + 12 + 4) + 6 + 4 + 1) = 115 possible distinct arrangements of those selections.

Hope that's right.

2007-01-21 15:10:02 · answer #1 · answered by Jim Burnell 6 · 0 0

decision: 12 association: 40 seven evaluate situations : a million) all are comparable : so all would be 'S' so no fo situations = a million arranegemt corresponing to this = a million 2) 3 comparable a million distinctive 3 S's and a million between P,O,E decision no of case = 3 association attainable = 4 3) 2 of one variety and a pair of of different variety 2 S's & 2 E's no of alternatives = a million association attainable = 4!/2!*2! = 6 4) 2 of one variety and relax differenr no of decision = 2C1*3C2= 6 association attainable = 4!/2! = 12 4) all distinctive decision = 4C4 = a million association = 4! = 24 finished decision attainable = 12 finished association attainable = 40 seven Assuming the taking part in cards with comparable letter are seen comparable.

2016-12-16 09:58:02 · answer #2 · answered by mundell 4 · 0 0

I dont know

2007-01-25 04:11:36 · answer #3 · answered by Brent S 1 · 0 2

fedest.com, questions and answers