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In general, the number of ways that "m" objects can be distributed into "n" boxes is n^m (When the objects and boxes are distinguishable, and a box can be empty) I've attached a little reference chart for you to remember the formula for these kind of problems. For the GMAT I wouldn't bother deriving them - just remember them.
So distributing the 7 boxes into the 7 days needs 7^7 ways.
It must be clear to you that if all 7 happened on the same day, it could happen in 7 ways, (since it depends on which day this happened)
Prob = 7 / 7^7 = 1 / 7^6
File comment: Here's a nice little chart for these kind of repetition problems.
Combinatorics.jpg [ 18.18 KiB | Viewed 4041 times ]
As I a remember there is one formula for distribution of n objects to r people, that formula consider all cases of distribution of the objects in chunk of 0 , 1 , 2 .. so on.
Formula = (n+r-1)C(r-1) (bcz i think accidents can happen or not happen on any particular day or even multiple accidents can happen) thus this formula qualifies to find all such cases.
13C6 = 1716
desirable cases are 7.
thus Prob = 7/1716
I hope i am right.
Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".
I don't understand the formula for counting the sample space. Your inputs/explanation would be very appreciated.
Pretty straightforward Cheers J
Can you please explain why 7C6 in numerator.
Sure. Each accident can happen in each of the 7 days. We want the scenario in which all happen in the same day Therefore, since all accidents will happen in the same day then we only need to pick in which day out of the 7 days will all the accidents happen
So 7C1 = 7C6 = 7
Hope its clear J
Re: If 7 distinct car accidents happened during 7 days, what is
31 Jan 2014, 13:03