Combinations and Permutations (1 Viewer)

[stupid idiot]

Originally posted by George W. Bush
Say we have no As, and we choose the letters R S and T.

According to your method, RST, RTS, TSR, TRS, SRT, STR are all the same word.

Bush, the question says combination of 3-letter words. So according to the question all the arrangements you listed are indistinguishable.

If i did count the permutations, ie different words that can be made, it came out to be something like 291 (i can't remember it now). But even that is no where near the required answer of 60480.

Any clue how this nike33 guy did the question and apparently got the right answer of 60480 i just could not get his/her logics.

 

[stupid idiot]

Someone's got to help me out, i'm having nightmares about combinatorics. Who would've thought high school combinatoric questions could be this tricky. I don't like spamming so please help.

HOW DO YOU GET 60480 WAYS

 

[-=«MÄLÅÇhïtÊ»=-]

Originally posted by stupid idiot
i do not get this at all. how come i got 42 ways for my answer?

# 3A's = 1
# 2A's = 6C1
# 1A's = 6C2
# 0A's = 6C3

# total ways = 1 + 6C1 + 6C2 + 6C3 = 42

what's wrong with my logics? did i not interpret the question right? fuck i'm stupid.

those are only combos, u have to permutate all the cases too. i'll do it now...

My solution:

This type of question is a typical one u solve by cases.
You have 3 cases, the words could be
3x (ie. all the same)
2x, y (2 the same, 1 different)
x,y,z (all 3 are different)

for each case u work out the combinations and then permutations.

first write out AUSTRALIA this way:
A:3, U:1, S:1, T:1, R:1, L:1, I:1

for the case of 3x:
Combinations:
1C1 (only 1 case where u cna have all 3 letters same)
Permutations:
3!/3!
Total:
1*1=1

2x,y:
Combos:
1C1 * 6C1 (1C1 because again A is only letter that can possibly appear twice, and 6C1 because u have 6 remaining letters and u wanna choose 1)
Perms:
3!/2!
Total:
6*3=18

x,y,z:
Combos:
7C3
Perms:
3!
Total: 210

Adding up all the totals from each case u get 229
(u prolly wanna set out the above in a table wiv columns of cases, combos, perms and totals, and have rows of the 3 cases.)

you obviously cant get the huge answer that previous person got because 9P3 would be the max answer without any restrictions. Look at his reasoning, u did 9P3 and then timed by something...the logic behind that is wrong...

 

[stupid idiot]

Ok ok, so nike33's reasoning is wrong? can you confirm that absolutely?

Because apparently he got the right answer which is quoted by the original question poser.

And one more thing which is not really important but just that you know the question did ask for "combination" which means it should be 42.

 

[nike33]

yes its wrong..ill attempt it agian

to pick any three letters there are 9p3 ways

let the a's be a1,a2,a3

so if there are 3 a's it would be a1,a2,a3 ..there are 6 ways of arranging these that the 9p3 took in account but we only ned one
hence (-5) from 9p3

next, if a1,a2 or a3 are selected

a1a2x (x is a remaining number..6 left hence 3p2.6 = 36)
a1xa2 (36)
xa1a2 (36)

but they are the same as a1=a2=a3 hence there are only 18 that need to be taken in account hence 36.3 = 108 then - 18 ie 90 as only 6 are diferent from each row

now (9p3 - 5 - 90)

now if only a1 or a2 or a3 taken (x first letter out of 6, y second)

a1,x,y (6.5 = 30 ways x3 replacing a1,a2,a3)
x,a1,y (similar)
x,y,a1 ('''')
so 30x3x3 ways but only 30 ways in each row are different ie only 90.

30x3x3 - 90 = 180

ans = 9p3 - 5 - 90 - 180

yay which is same as malcalite