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# In how many ways can a person draw two cards at random from

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VP
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In how many ways can a person draw two cards at random from [#permalink]

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04 Jan 2006, 02:30
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In how many ways can a person draw two cards at random from a pack of 52 such that the two cards belong to two different suits? For e.g, if one card is a diamond the other should belong to one of the other three suites?

A) 13*13
B) 52*39
C) 13*13*4
D) 13*13*4C2
E) 52C2-13C2
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04 Jan 2006, 03:08
I think its D. 13*13*4C2

Take two suites at a time out of four.
Number of ways two suites can be chosen from 4 = 4C2

Out of these two suites number of ways we can have different cards = 13*13.

So Ans = 13*13*4C2
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Last edited by ps_dahiya on 04 Jan 2006, 12:58, edited 1 time in total.

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04 Jan 2006, 05:10
2 out of 52 in 52C2=1326 ways
2out of 13 in 13C2=78 ways so requred number is E)

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04 Jan 2006, 05:51
Card one: 52 ways to pick

After picking card one, we need to discard the remainig from the suite - 13 cards

Second card would therefore have 52-13 possibilites = 39

# of ways = 52 * 39

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04 Jan 2006, 05:51
@ BG

According to your logic you need to multiply 13C2 with 4, since there are four suits a 13 cards.

======================

@ ywilfired

I can't evaluate why 13*13*4C2 should be wrong.

13*13 means the 13^2 ways to combine the cards two suits.
4C2 means the number of combinations you can choose two suits out of four.

What is wrong with it?

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04 Jan 2006, 09:50
4 * (13C1 * 39C1) = 4*13*39

Answer B ... 52 * 39

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04 Jan 2006, 13:42
@ BG

According to your logic you need to multiply 13C2 with 4, since there are four suits a 13 cards.

======================

@ ywilfired

I can't evaluate why 13*13*4C2 should be wrong.

13*13 means the 13^2 ways to combine the cards two suits.
4C2 means the number of combinations you can choose two suits out of four.

What is wrong with it?

Allow me to clarify. What I have learnt here is

1. If question asks # of ways then the order matters.

i.e 52*39

2. If question asks how many different combinations then order doesn't matter.

i.e 13C2 * 13C2 * 4C2

Lets prove this by taking 8 cards of four suites. Say A1, A2, B1, B2,C1,C2 and D1,D2

For case 1 above:
(A1, another card other than A2) = 6 ways
(A2, another card other than A1) = 6 ways
(B1, another card other than A1,A2,B2) = 4 ways
(B2, another card other than A1,A2,B1) = 4 ways
(C1, another card other than A1,A2,B1,B2,C2) = 2 ways
(C2, another card other than A1,A2,B1,B2,C1) = 2 ways

Total = 24. But there is difference between (A1,B1 - First A1 drawn and then B1 drawn) and (B1.A1 - First B1 drawn and then A1 drawn) if # of ways are asked.

So total = 24 * 2 = 48 i.e 8* 6 (Equivalent to 52*39)

For case 2 above:
(A1, another card other than A2) = 6 ways
(A2, another card other than A1) = 6 ways
(B1, another card other than A1,A2,B2) = 4 ways
(B2, another card other than A1,A2,B1) = 4 ways
(C1, another card other than A1,A2,B1,B2,C2) = 2 ways
(C2, another card other than A1,A2,B1,B2,C1) = 2 ways

Total = 24. There is no difference between (A1,B1) and (B1.A1) if # of diferent combinations are asked.

So total = 24 i.e 2 * 2 * 4C2 (Equivalent to 13 * 13 * 4C2)

Hope this clears all doubts. I also calculated wrong but good thing is I learned a new concept.
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05 Jan 2006, 03:53
Thanks dahiya

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05 Jan 2006, 03:53
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