HappyMathtutor wrote:
In case pure combinatorics is too tough, this is a bit of a longer way, but it explains the how and why
there are only 3 possibilities where the range can be 7.
(1,8) (2,9) (3,10)
If we choose (1,8),
the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9
now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168
But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.
12*1/168= 1/14
If we choose (2,9)
the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9
now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
If we choose (3,10)
the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.
now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,
so the next fraction is 6/8.
now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7
(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14
there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B
1/90+1/120+1/168= 8/315
Any pointers on how to bridge this gap?
4C2 is 6. How come it is 12?