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23 Jul 2025, 06:00
Kudos
Bookmarks
Hi, I am a little confused here. Help me with where I am going wrong and why.
I divided this problem into 2 parts...
in which the probability of getting 3 sixes and 2 different numbers...
9 x 8 x 5c3 = 1440
and probability of getting 3 sixes 2 of the same numbers
(9 x 1 x 5c3) / (3! x 2!) = 90
Total choices = 10^5
Adding both and dividing by total choices, I get (E)
I am not sure why the original solution has not taken these cases. Can you please explain?
I divided this problem into 2 parts...
in which the probability of getting 3 sixes and 2 different numbers...
9 x 8 x 5c3 = 1440
and probability of getting 3 sixes 2 of the same numbers
(9 x 1 x 5c3) / (3! x 2!) = 90
Total choices = 10^5
Adding both and dividing by total choices, I get (E)
I am not sure why the original solution has not taken these cases. Can you please explain?
23 Jul 2025, 14:09
Kudos
Bookmarks
Nipunh
9 * 8 * 5C3 = 720, not 1440.
For the second case, it's just 9 * 5C3 = 90.
Total is 810.
However, the official solution uses 9 * 9, which counts all combinations of the two remaining digits, including both same and different.
10 Sep 2025, 12:35
Kudos
Bookmarks
During the test i got it wrong..but later while reviewing ...this was my approach
Total cases can be 10^5
Now...Favourable....5 spaces are there.....
The number 6 can choose any 3....So out of the 5 ...6 can choose its place in 5C3 Ways....Makes its 10...And for each place we have 9 X 9 options ..So 81 codes for each of the 10 ways....
Makes it 810....
Answer 810/10^5
Total cases can be 10^5
Now...Favourable....5 spaces are there.....
The number 6 can choose any 3....So out of the 5 ...6 can choose its place in 5C3 Ways....Makes its 10...And for each place we have 9 X 9 options ..So 81 codes for each of the 10 ways....
Makes it 810....
Answer 810/10^5
