In combinations it does not.So, the permutations 1,2,3 and 3,1,2 are not the same. In permutations the order of the elements does matter. What is an arrangement in which order is important A permutation is the correct answer. If we regard these N-r objects left behind as the chosen objects, we see that the number of combinations of N objects taken r at a time is exactly equal to. The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is. You would use combination when order does not matter. Permutations When Order Matters By: Mary Jane Sterling Updated: 03-26-2016 Linear Algebra For Dummies Explore Book Buy On Amazon Permutations involve taking a specific number of items from an available group or set and seeing how many different ways the items can be selected and then arranged. Just think about how many different ways you can arrange A and B. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$). Remember: order matters As an example, consider the permutations of the letters A and B. Permutations: order matters, repetitions are not allowed. This is '(20) choose (3)', the number of sets of 3 where order does not matter. Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. For combinations, we chose (3) people out of (20) to get an A for the course so order does not matter. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ( $n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s). Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Notice that we can characterize the sum this way: $$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$ If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see A true 'combination lock' would accept both 10-17-23 and 23-17-10 as correct. The order you put the numbers in matters. You know, a 'combination lock' should really be called a 'permutation lock'. The same logic applies in the general case but it becomes murkier through the abstraction. Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter). The reason combinations come in can be seen in using a special example.
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