selection

[sunilrawat]

I have a doubt.
Suppose we have to select 2 people from a group of 5, we will do it in 5C2 ways.
But if the selection is to be made such that two specific people are never selected, as per my understanding, we will have to divide 5C2 by 2!. Am I right?

[user123321]

I will try to explain

suppose A,B,C,D,E are five people...if two specific people say A,B is to be excluded, then from remaining three people, two can be selected in 3C2 ways.

user123321

[sunilrawat]

I think I mixed up my query.
Let me put the exact question.

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33

I tried solving it with two methods (one wrong), spent a long time, and got the answer right eventually. Here is the method I got wrong.

First I considered the total possible outcomes: 12C4
Then I calculated the favourable outcomes as follows:
The first card can be be selected from 12 cards in 12C1 ways
The second card has to be selected from remaining 10 cards, (excluding the pair of already selected card) = 10C1
Similarly, for third card = 8C1
and for fourth, = 6C1
So the probability = 1-((12*10*8*6)/12C4), which is the wrong answer.
I figured that while selecting the second card, I was also including the possibility of selecting the paired card as there were still 11 cards in the deck.

Then I applied following method:
Probabality of first card = 12/12 = 1
Probabality of first card = 10/11
Probabality of first card = 8/10
Probabality of first card = 6/9
Thus, required probability = 1-((1)*(10/11)*(8/10)*(6/9))=17/33, the right answer !

This means if I divide the expression (12*10*8*6) from my first method by 4!, I get the right answer.
What I am not able to understand however is that what exactly am I doing by dividing by 4! ?
Please explain with an example.

[knight247]

Sunil,
What u've done in first method where u've got the answer wrong, is u've used the slot method. When u use the slot method, you always get the ordered sequences i.e. the permutations which are not required in this case. Thats why u need to divide it by 4! i.e. 24 inorder to convert the ordered sequences to unordered sequences i.e. combinations. You've actually figured that part by urself.lol. Cheers