Count The Number of Arrangements

There are 10 parking spaces numbered from 101 to 110. At least one car is parked in these slots. If cars can be parked only at the consecutively numbered parking slots, how many such arrangements can be made. 

Consider that only one car can be parked in one parking slot and all cars are identical.

Here is the possible count! 

Possible Number of Arrangements

What was the puzzle?

Suppose there is only 1 car that is to be parked in 1 of the 10 slots. 

Number of possible arrangement = 10C1 = 10!/1!9 = 10.

That is 1 car can be parked in 10 slots in 10 number of ways.

Now, let's suppose that there are 2 cars that to be parked in 2 of the 10 parking slots. But the condition is that they need to be parked in consecutive slots. 

Among 10 slots for there are 9 possible consecutive slots for 2 cars. That is, 2 cars can be parked in consecutive slots in 9C1 = 9 number of ways. It's like placing 1 group of cars (having 2 cars) in 9 possible slots.

Similarly, in 10 parking slots for parking 3 cars there are 8 possible consecutive slots. Hence, there are 8 such arrangements are possible.

And so on for the rest number of cars.

Hence, there are total 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 such arrangements are possible.  

Permutations or Combinations or Common Sense?

You have 12 black socks and 12 white socks mixed up in a drawer. You’re up very early and it’s too dark to tell them apart. What’s the smallest number of socks you need to take out (blindly) to be sure of having a matching pair?

To know what it is click here! 

Source 

It is Common Sense not Permutaions nor Combinations!

Question is here!

You need only 3 attempts.

Let's say on first draw we get BLACK socks. In next draw, if we get BLACK then we have matching pair immediately. But if in second draw, it's WHITE then we have to go for third draw. 

On a third draw, if it's BLACK then we have pair of BLACK socks ; one came from first draw & other from third one!

On the other hand, if we get WHITE on third draw then we have pair of WHITE socks ; one we got from second & other from third.

Same logic can be applied if WHITE comes in first draw.