Problem Solving

amisahoo wrote:A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected from the deck, what is the probability that the numbers on the cards are drawn in order ?

A. 1/60
B. 1/216
C. 1/30
D. 1/24
E. 1/6

Please help me out with this problem . Some people take it as a arrangement problem but here we are doing selection and not arrangement.

This is one of those terribly worded prep company questions that is open to multiple interpretations. When I read the phrase 'the cards are drawn in order', I have no idea just how to interpret that. If I pick, in order, 2, 3 and 5, those are in increasing order - does that count? And if I pick 6, 2, 1, those are in decreasing order, so also 'in order' - does that count? If they mean to ask about selecting consecutive integers (and from the OA, I can only guess that's what they meant), they need to say that, and even then, would 4,3,2 be a legitimate selection, or do the numbers need to be in increasing order?

If we want to know the probability that we pick three *consecutive* numbers in *increasing order*, there are 6*5*4 possible sequences we can pick in total, so that's our denominator, and there are 4 ways to pick an increasing sequence of consecutive integers (we can pick 123, 234, 345 or 456). So the answer is 4/(6*5*4) = 1/30.

If instead you were asked (this is how I first interpreted the question you posted, until I saw what they think is the OA) 'If you pick three cards one at a time and without replacement from a deck of six cards numbered from 1 to 6, inclusive, what is the probability the numbers picked are in increasing order', the probability would be higher, because there are many increasing sequences we could pick which do not consist of consecutive numbers. Here the answer would be 1/6. One way to look at this question: for any three numbers I could pick, there are 3! = 6 orders I could pick them in. Only 1 of those orders is increasing, so the probability I selected my numbers in that particular order must be 1/6. Notice that the number of cards in the deck is completely irrelevant in this question; we could have a deck of 100 cards numbered from 1 to 100 and the answer would be the same.