## Problem Of The Week 31

**A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park? **

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Let the parking spaces be 1,2,3..16. Now if at least two adjacent spaces are empty, the number of ways are choosing are 15((1,2)(2,3)…(15,16))..Now from the rest 14, we can choose 2 empty spaces(So this would cover the other cases of 3 adjacent spaces and 4 adjacent spaces too). So the total number of ways would be 15*14c2 and hence the probability is 15*14c2/16c4 = 3/4

Celebrating LifeSeptember 29, 2008 at 8:18 am

look again bhaskar

this one is wrong

outtimedSeptember 29, 2008 at 11:10 am

It is actually easier to calculate the probability that empty spaces are all not mutually adjacent and take 1 minus that probability which gives us the probability that there are at least 2 adjacent empty spaces. The number of non mutually adjacent empty spaces is 13c4. It can be viewed as inserting spaces in between the 12 cars; 13 possible places to insert 4 places. So the probability that she can park is 1 – 13c4/ 16c4.

This question appeared in one of the AMS competitions. Should be easy to google for.

Christopher TaySeptember 29, 2008 at 1:18 pm

yupp, taken from there !!

I felt this was a nice practice problem!

outtimedSeptember 29, 2008 at 1:24 pm

I am not able to understand Tay’s solution. Please explain it.

QuarkSeptember 30, 2008 at 5:53 pm

too gud to solve………haha

wolverineOctober 1, 2008 at 5:43 pm