Combinations and permutations question.

[delboi47]

Hello all,

I have a choice of using the following counting methods to answer the questions: multiplication, permutation equation P(n,k) = n!/(n-k)!n!, combination equation C(n,k) = n!/(n-k)! and P(event occurs) = number of desired events/ total possible events.

A question reads:
Given a standard deck of playing cards, what is the probability of being dealt the following cards (in the order stated):

c. Two aces and two jacks

d. An ace, the jack of hearts and another jack

Working so far for c:
The issue I have here is the "in the order stated" part of the question. To me that suggests using the permutation equation, however it doesn't specify the exactly what ace or jack to have which lead me to use the combinations equation.


1) My working out is as follows:
Aces: 4C2 = 6
Jack: 4C2 = 6
desired events = 6*6 = 36.

total possible events = 52C4 = 270725 (here I assumed that because there are a total of 4 cards being drawn in the question that this is all the possible outcomes of them)

Answer = 36/270725 = 1.33x10^-4

2) Another way I attempted was using the multiplication method but eventually got stuck.
4/52*3/51*4/50*3/49 (with respect to 2 Aces then 2 Jacks)

Thanks in advance for your help!

Working out for d:
I haven't had a proper attempt at this question as I have been stuck on C. However, I believe that the Jack of hearts becomes a 1C1 combination followed by a 3C1 for the other jack. OR 1/51*3/50.

 

[Harry_the_cat]

The second way you have done (c) is correct. Where are you stuck? Answer is 144/649400. Just multiply what you have.
Your first attempt as (c) doesn't take ORDER into account.

For (d):
What is the prob that you deal an ace first?
What is the prob that you then deal the jack of hearts (given that the first was an ace)?
What is the prob that you then deal another jack (given that an ace and the jack of hearts have been dealt)?

You are on the right track above. I'd forget about the C notation altogether and just use multiplication of probabilities.

 

[delboi47]

The second way you have done (c) is correct. Where are you stuck? Answer is 144/649400. Just multiply what you have.
Your first attempt as (c) doesn't take ORDER into account.

For (d):
What is the prob that you deal an ace first?
What is the prob that you then deal the jack of hearts (given that the first was an ace)?
What is the prob that you then deal another jack (given that an ace and the jack of hearts have been dealt)?

You are on the right track above. I'd forget about the C notation altogether and just use multiplication of probabilities.

For C I got stuck assuming that I had to divide what I already worked out by the total possibilities. e.g. 52! or 52C4 as this is what I found out from some what similar questions on the web.

Thanks for your response! I think I may have been over complicating things...