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by shankar.ashwin » Sat Sep 24, 2011 2:10 am
There are 10 coins numbered from 1 to 10 in a box. In how many ways can I pick 4 coins one after the other, such that, each time I get a coin numbered which is greater than what I picked up the previous time?

A) 6
B) 210
C) 840
D) 2520
E) 5040
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Source: — Problem Solving |

by cans » Sat Sep 24, 2011 3:48 am
10C4.
(select any 4 numbers. way of arranging them=1)
10C4 = 10*9*8*7/24 = 210
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by knight247 » Sat Sep 24, 2011 12:20 pm
For each coin that is picked to be greater than the previous one, the basic requirement is that each of the four coins be different. So lets pick 4 coins out of 10 and arrange them in 10P4=10*9*8*7=
5040 ways.

Now, lets consider a random scenario of different numbers 1 2 3 4 which can be arranged amongst themselves in 4!=24 ways. But out of these 24 ways only 1 way i.e. 1234 is our desired outcome. So, 1 out of every 24 outcomes will be our desired outcomes. So (1/24)*5040=210 Hence B
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