Probability question (1 Viewer)

[wiz g 99]

A box contains six cards, two of which are identical. From this box, three cards are drawn in order without replacement and put on a table in a row. How many different arrangements can be made?

The answer is 72. I've answered 6P3-4C1*3P3*2P2 to eliminate the repetitions, but 48 repetitions seems too much. Correct me if I'm wrong but 48 repetitions means that 96/120 of the original with-repetitions arrangements contain both identical cards.

 

[snowflakeptical]

hey can you do up a probability tree?
6 cards- I labelled A B C D E(x2)
A= 2 x 3 x 4 (times that by 5 instead of 6)
B
C
D
E
(only did one e as the other is repeating) so I got 120 but I'm not sure how your probability tree turned out
better be safe by drawing all the results


*wait I realised my mistake. E has a special exception. I suggest drawing up all the possibilities

 

[InteGrand]

A box contains six cards, two of which are identical. From this box, three cards are drawn in order without replacement and put on a table in a row. How many different arrangements can be made?

The answer is 72. I've answered 6P3-4C1*3P3*2P2 to eliminate the repetitions, but 48 repetitions seems too much. Correct me if I'm wrong but 48 repetitions means that 96/120 of the original with-repetitions arrangements contain both identical cards.

Let the cards be X, X, A, B, C, D, so X and X are identical cards.

There are three cases: have no X's in the three picked, have exactly one X, or have exactly two X's.

If you have no X's, you're making an ordered trio from A, B, C, D, and number of ways to do this is ⁴P₃ = 24.

If you have exactly one X, you need to pick two cards from A, B, C, D (which can be done in ⁴C₂ = 6 ways), and then order X and the other two cards (which can be done is 3! = 6 ways), so total no. of ways if you have exactly one X is 6×6 = 36.

If you have two X's, you pick the last card from A, B, C, D (4 ways to do this), then order the two X's and the last letter (3 ways to do this), so 4×3 = 12 ways for this case.

So total no. of ways is 24 + 36 + 12 = 72.

 

[Kaido]

yeah, the lesson is to never jump into permutation/comb questions without a game plan

How did you get "6P3-4C1*3P3*2P2"??

 

[dan964]

Consider A, B, C, D, E, E

Possibility one, I pick 3 out of A, B, C, D = 4P3 = 24
Possibility two, I pick 2 out of A, B, C, D and an E
= 4P2*3P1 (the 3P1 here is to account for the E)
(or 4C2 * 3!)
=36
Possibility three, I pick one out of A, B, C, D and both Es.
= 4P1*3P1 (the 3P1 here is the placement of the non E)
=12


Total = 72 which is the correct answer

 

[Kaido]

You choose 2 of the four then permutate it 3! times.
likewise for the last last part: you choose 1 of four then permutate it 3! times and divide it by 2! due to repetition

 

[dan964]

^
what if you pick 2 that are identical. you didn't consider that case.
I have amended my answer above.
and no I didn't look at Integrand's answer - couldn't be bothered scrolling up.

 

[Kaido]

you choose 1 of four then permutate it 3! times and divide it by 2! due to repetition

this refers to the two identical case (since you choose 1 of A,B,C,D and 2 E's)

but you got it anyways >.>

 

[wiz g 99]

yeah, the lesson is to never jump into permutation/comb questions without a game plan

How did you get "6P3-4C1*3P3*2P2"??

Hmm, I thought the way you solve it would be total-repetitions and somehow I got 6P3-4C1*3P3*2P2.

 

[braintic]

permutate it 3! times

The verb is 'permute'.

 

[Kaido]

lol no wonder, had a red line under it while i was typing

still, permutate sounds pro

 

[braintic]

lol no wonder, had a red line under it while i was typing

still, permutate sounds pro

"Permutate" is an example of a 'back-formation".

Another example is "coronate". I wonder how many people think it is correct to say "Queen Elizabeth was coronated in 1953" ?
Then there is fragmentate and spectate.

However there are many back-formations that have (unfortunately) become acceptable - like commentate, elocute, notate, orientate and surveil.