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- Thanks a lot! The edit portion is so insightful.
- Thanks! Will need to go through in order to understand.
- In a way, each way created on the numerator with 12! ways will have 4 going into team 1, 4 into team 2, and remaining 4 into team 3. It can be intuitively visualized as 3 blocks with each block containing 4 elements. The denominator then takes care of overcounting.
- Yes, the answers help clear the doubts.
- Thanks!
- Thanks!
- What makes my solution wrong is perhaps I am ruling out the possible permutations and combinations that each can have with the remaining 11. So I am just counting one or partial instances of the total possibilities. When 12! placed on numerator, all the permutations and combinations taken care of. It will help to have […]
- How many ways 12 people be divided into 3 teams with 4 people each. My argument was 4! x 4! x 4! x 3!. But the correct answer is 12!/( 4! x 4! x 4! x 3!). It will help to have an explanation why my argument is incorrect. Update 1: What makes my solution […]
- For exact stats, GA and Jetpack (for WordPress) are more reliable. Tools like Semrush and Ahrefs are approximations. They usually show 0 or no organic traffic for starter, smaller sites.
- https://www.canva.com/design/DAG461SWfV0/T6ZLlHfPeppvqFQAAepJ8w/edit?utm_content=DAG461SWfV0&utm_campaign=designshare&utm_medium=link2&utm_source=sharebutton The problem is finding probability how many 26 letter unique words can be formed with 26 letters in English alphabet against the words of different lengths (that is summing no of ways one letter word can be formed, two letter words can be formed and so on till 26). Here is how I tried: […]
Last Updated on October 7, 2025 by Admin
