Possible Combinations from Cards Drawn at the Same Time?Date: 04/02/98 at 22:40:48 From: Gladys Subject: Probability Four cards were placed in a bag, each having a different number written on it. All of the cards were the same size, shape and color. If two of the four cards were drawn AT THE SAME TIME out of the bag, how many possible combinations could be made? a. 4 b. 6 c. 12 d. 16 Our class thinks the answer is b (6), but our teacher's book says the answer is c (12). We think the teacher's book is wrong because they "forgot" the words "at the same time." Can you help us? Also, is there a formula for solving probability problems? Thank you, Gladys from Mrs. Allen's class Date: 04/03/98 at 07:30:52 From: Doctor Anthony Subject: Re: Probability Two objects can be chosen from 4 objects in 6 ways, but the question then asks how many combinations of numbers could be made. This is a poorly worded question, because you could interpret the last part in two ways: namely, the number 5,3 (for example) might be thought different from 3,5, or you might think these are the same combination. My preference is the answer you gave, 6. There are general formulae for these types of problem. You will find them in textbooks under the headings "Permutations and Combinations." If you have, say, 10 objects, then the number of arrangements (permutations) that could be made from 4 of the objects is given by: P(10,4) = 10*9*8*7 = 5040 This is because you could choose the first object in 10 ways, the second object in 9 ways, the third in 8 ways and the fourth in 7 ways. In this question, the ORDER of the objects matters, so that, for example, swapping the first and second counts as a different permutation. Using the factorial notation where say 5! = 5*4*3*2*1 = 120, we could write: 10! P(10,4) = ----- = 5040 6! With COMBINATIONS, we are not concerned with the order of the objects. The number of combinations of 4 things that could be chosen from 10 things is given by: 10*9*8*7 P(10,4) 10! C(10,4) = ---------- = -------- = ------ 4*3*2*1 4! 4! 6! This is because each of the PERMUTATIONS of 4 things could be rearranged within itself in 4! ways without counting as a different COMBINATION. So the expression for the number of permutations is too large by a factor of 4! when we are trying to find the number of combinations. This means we have the formula for C(10,4) as: 10! C(10,4) = ------- 4! 6! The question you started with was how many ways can we can choose 2 objects from 4. 4! C(4,2) = ------ = 6 2! 2! -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/