An urn has 10 red cards numbered 1 through 10 and 8 blue cards numbered 1 through 8. Three cards are randomly drawn, one at a time, without replacement.

Find the number of ways to obtain:

A) all red cards.

B) all blue cards.

C) all same colour cards.

D) 2 reds and 1 blue.

E) exactly one card numbered 1.

F) two cards numbered 1.


Seeing as we pick one card at a time, I assuming that order is involved which means that this a permutation problem.

Parts A and B are straightforward: $_{10}P_{3}$ would be for Part A and $_{8}P_{3}$ would be for Part B.

Part C is simply: $_{10}P_{3}$ + $_{8}P_{3}$

Part D to F is where I'm having trouble.

For part D, what I think the answer is: ($_{10}C_{2}$ + $_{8}C_{1}$) * 3!

My thought process was to find the combination of getting two reds and one blue and multiplying by 3! because that is how many ways you can arrange it.

  • $\begingroup$ How about $_{10}C_{2} \times _{8}C_{1}$? $\endgroup$
    – zed111
    Jul 11, 2015 at 19:34
  • $\begingroup$ @zed111 Can you explain why? $\endgroup$
    – Deathslice
    Jul 11, 2015 at 19:36
  • $\begingroup$ See rule of product en.wikipedia.org/wiki/Rule_of_product $\endgroup$
    – zed111
    Jul 11, 2015 at 20:00
  • $\begingroup$ I'm doing the same problem, but I don't understand the first part. In A, wouldn't that expression give you the redundant values of equally valued groups of 3? like ABC has the same value as BAC if we talk about cards? meaning that this are not permutations? $\endgroup$
    – dacabdi
    Dec 10, 2015 at 20:55

2 Answers 2


D) There are $_{10}C_2$ ways of choosing $2$ cards from $10$ red cards and $_8C_1$ ways of choosing 1 card from 8 blue cards. So the total possibilities are : $_{10}C_{2} \times _{8}C_{1} \times 3!$

E) The card numbered one can be chosen in $2$ ways. Then the remaining two can be chosen in $_{10+8-2}C_2$ ways. So the possibilities are: $2\times_{16}C_2\times3!$

F) Two cards numbered one can be chosen in 1 way. Then the other card can be chosen in $_{16}C_1$ ways. Total possibilities are: $_{16}C_1\times 3!$

  • $\begingroup$ In part F, the other card can be chosen in $C(16, \color{red}{1})$ way since only one other card is being selected. $\endgroup$ Jul 12, 2015 at 10:35
  • $\begingroup$ Oh yes. Thanks. $\endgroup$
    – zed111
    Jul 12, 2015 at 10:43

You're close. What you want is $({}_{10}C_2 \cdot {}_8 C_1) \cdot 3!$, not $({}_{10}C_2 + {}_8 C_1) \cdot 3!$.

What's the difference? Multiplication counts the ways of doing BOTH of two things. Addition counts the ways of doing ONE OR THE OTHER of two things. In your case, you need to pick $2$ of the $10$ red cards AND $1$ of the $8$ blue cards, because you need $3$ cards total.

  • $\begingroup$ but aren't they drawn one at time, not in succession? $\endgroup$
    – Deathslice
    Jul 11, 2015 at 19:58
  • $\begingroup$ Never minded, somebody explained for me. Thanks for your help. $\endgroup$
    – Deathslice
    Jul 11, 2015 at 20:06

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