part 2 of commenting on all the videos in this series as you deserve it for your hardwork. Not only me but even random people I meet who happen to take discrete math 1 and 2 just coincidentally during the conversation bring you up because you are the universal study guide for discrete math. What an amazing teacher you are.
Thank you for putting this out there. This course is a wonderful overview of the subject. I've gone through the material in my course and it felt disjointed. Nice to piece it together with your playlist.
I thought the same thing. I was thinking 5! You can think of it like a four of a kind hand in poker. You have 13 ways of doing four of a kind, and then there are 48 other cards to complete that hand, so it’s 13 * 48. Here the four of a kind is like the “ABC” and it’s basically held constant. In this question the “kind” (i.e., “ABC” has only one instance), so I thought the answer is 5! * 1.
The correct answer is 6! and I hope I can help clarify why :). I thought it was 5! at first because I misunderstood the question. If ABC had to be the FIRST 3 letters, the solution would be 5!, because we would know that the first position always has to be ABC. However, ABC doesn't need to be at the beginning and instead it can be anywhere in the string (i.e. DABCEFGH would still count). Because of that, we treat ABC as one "letter", since they all need to be together, but it's not 1 * 5! because it's not guaranteed to be in the first position of the string. Instead it can be 1st, 2nd, 3rd, 4th, 5th, or 6th in the string. Basically, we have 6 "letters" that can go in any order, which is why 6! is the solution.
part 2 of commenting on all the videos in this series as you deserve it for your hardwork. Not only me but even random people I meet who happen to take discrete math 1 and 2 just coincidentally during the conversation bring you up because you are the universal study guide for discrete math. What an amazing teacher you are.
Thank you for putting this out there. This course is a wonderful overview of the subject. I've gone through the material in my course and it felt disjointed. Nice to piece it together with your playlist.
thank you so much for these series, theyre all complete and utter life savers
Welcome back mam ☺️
On the last question isn’t it supposed to be P(5,5)?
Since n=6 & one of the elements stays the same ?
I thought the same thing. I was thinking 5! You can think of it like a four of a kind hand in poker. You have 13 ways of doing four of a kind, and then there are 48 other cards to complete that hand, so it’s 13 * 48. Here the four of a kind is like the “ABC” and it’s basically held constant. In this question the “kind” (i.e., “ABC” has only one instance), so I thought the answer is 5! * 1.
yup ur right. she made a minor mistake. 👍
The correct answer is 6! and I hope I can help clarify why :).
I thought it was 5! at first because I misunderstood the question. If ABC had to be the FIRST 3 letters, the solution would be 5!, because we would know that the first position always has to be ABC. However, ABC doesn't need to be at the beginning and instead it can be anywhere in the string (i.e. DABCEFGH would still count).
Because of that, we treat ABC as one "letter", since they all need to be together, but it's not 1 * 5! because it's not guaranteed to be in the first position of the string. Instead it can be 1st, 2nd, 3rd, 4th, 5th, or 6th in the string. Basically, we have 6 "letters" that can go in any order, which is why 6! is the solution.
@@melodyjadynThanks for explaining!!
Thanks a ton mate@@melodyjadyn
Amazing, love your videos
Thank you so much!!!
👍