# Singles – 28

## 8 thoughts on “Singles – 28”

1. kishor |

in 2 question in 2 step the set should be 31

• No, since the case (20, 20, 20) has already been accounted for.

regards
J

2. Tanya |

Octal number system has 8 digits from 0 to 7. How many 4 digit numbers in the octal number system exist such that they are formed using exactly three distinct digits?

here can we do 7*7*6*6 (7, for digits 1-7, and then 7, since we include 0, and so on) =1764 (which is the right answer) but i am not sure if this method is right, as the solution had a lot of cases, so what am i interpreting wrong here?

• So, this is not a place to discuss Simcat questions, please note ðŸ™‚ I am not sure what logic you have used (could not make sense of the numbers you have taken, as you haven’t accounted for the repeated digit I felt), but if you can justify/prove it then by all means use it.

regards
J

3. Vivek |

Sir, Can you please explain, how for finding total possible solution, you have used “18C2”?
Thanks!

• As mentioned, using partitioning. I am assuming a knowledge of basic PnC…

regards
J

• Vivek |

Sir,
sorry but I am still confused over “18”. I am imagining , a 16 m long chain, made using 16 numbers of 1m long smaller chains. We can make a cut at 17 places to break this chain into 3 units, so we can make any 2 cuts out of 17, but that gives “17C2”. I am not getting what is wrong in this way of thinking.
Thanks.

• Vivek |

Sorry sir, I got it, if in the above example, we make first cut at “0” position, we have 17 ways for next cut, then if first is at “1” position we have 16 ways, and so on. So number of ways is 17+16+15+……1 = (18*17)/2. Which is same as 18C2