dave13 wrote:
VeritasKarishma wrote:
dave13 wrote:
VeritasKarishma thanks, is it possible that you use my method for solving question ?
That method involves multiple levels of calculations which makes it cumbersome at test time.
A pair and two Reds:
6 * 5C2 = 6 * 10 = 60
A pair and two Blacks:
6 * 5C2 = 6 * 10 = 60
A pair and a red and a black non pair:
6 * 5 * 4 = 120
5 ways to pick a red and 4 ways to pick a non matching black.
(All I have done in my method above is combined these three steps together to get 240)
Two pairs:
6C2 = 15
Total = 255
many thanks
VeritasKarishma , i almost understand it
just have a few questions to make it clear for me
as per you method (see highlighted portion below):
"Select 1 pair out of the 6 in 6C1 ways = 6 ways (say you selected (R1, B1))
Of the leftover 10 cards, select any two in 10C2 = 45 ways.
But of these 45 ways, 5 are such that you have another pair (R2, B2), (R3, B3) ... (R6, B6)So you have 40 ways of selecting exactly 1 pair.
Total = 6 * 40 = 240 ways"
you say that out of 10 cards, you still can have pairs so to avoid another same pair you exclude 5 cards (R2, B2), (R3, B3) ... (R6, B6)
and you are left with 40 ways ... this is how i understand it ...though a bit confusing
but when i wrote this:
" - 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same) "
here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know
Btw, here "A pair and a red and a black non pair: 6 * 5 * 4 = 120" if third card is red which is fifth, in the rest of 4 cards there is a card of identical value among black ones left, so my question is why you didnt exclude that black card .... i mean i would do it 6*5*3 .... thats kinda confuses me
Yes, out of 10 when you pick 2 cards, they can be anything (R2, B3), (R4, R5), (B2, B6)... etc. You get 45 such cases. But 5 of these cases will be (R2, B2), (R3, B3), (R4, B4), (R5, B5), (R6, B6)
which give us another pair. So you need to remove them.
Quote:
here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know
I don't understand what you mean by - "I excluded 4".
If you want 1 pair and 1 red, 1 black of different values, you will select these as:
6C1 (6 ways to select a pair. Say you got R1, B1)
Now there are 5 ways to select R (R2, R3, R4, R5, R6). Say you got R2.
Now there are 4 ways to select B (B3, B4, B5, B6) to ensure that you get different values. You have already removed B2 to ensure that you do not get a pair. You are still left with 4 different values of B.
So it will be 6 * 5 * 4
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Karishma
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