![]() ![]() With $3$ people and $3$ awards, there would be $3^3=27$ ways to give these awards to the people. If we can give multiple awards to each person, we have replacement. ![]() Applied to your problem, imagine a variation in which instead of ranks we were giving $A,B,$ and $C$ awards. But since we put it back, on the next draw we also have $3$ possibilities! Thus, if we draw $k$ times, we have $3^k$ ways to pull the names. Commonly used mental models for this include 'urns in balls', 'balls in boxes', etc. For the first name we pull, we have $3$ possible names to pull. 129 1 7 Have you heard of the twelvefold way It is best understood in terms of counting maps (arbitrary, injective or surjective) up to equivalence (either just identity, or permutation of domain, or permutation of codomain, or both). How many ways are there to pull the names? Say we have $3$ names. This occurs when you allow replacement of the objects! Imagine you're pulling names from a hat, but every time you pull a name and read it, you put it back. The number of permutations of n objects, without repetition, is Pn P n n: The counting problem is the same as putting n distinct balls into n distinct boxes, or to count bijections from a set of n distinct elements to a set of n distinct elements. Understand the Permutations and Combinations Formulas with Derivation, Examples, and FAQs. Permutations are arrangements of objects (with or without repetition), order does matter. Permutations are understood as arrangements and combinations are understood as selections. It is very important to make the distinction between permutations and com- binations. After this issue, my approach is to multiply this '10' value into the possible placements for the remaining people, which Id expect to be 4, given that there are four people and four spaces in which to put them, so as with the full arrangement list, this would be 4/(4-4) 24. The primary distinction between permutation and combination is how the items or variables are arranged. The $3^3=27$ answer comes in a slightly different variation which involves neither permutations or combinations. Permutation and combination are the methods employed in counting how many outcomes are possible in various situations. Using a permutation of n2 also does not work, and does not seem logical. To arrange $n$ items in a row (which can be accomplished in $n!$ ways) is equivalent to picking $k$ of $n$ items to arrange in a row (which can be accomplished in $\frac=3!$ - exactly the result we were looking for! ![]()
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