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To refer to combinations in which repetition is allowed, the terms k-combination with repetition, k-multiset, [2] or k-selection, [3] are often used. [4] If, in the above example, it were possible to have two of any one kind of fruit there would be 3 more 2-selections: one with two apples, one with two oranges, and one with two pears.
In this example, the rule says: multiply 3 by 2, getting 6. The sets {A, B, C} and {X, Y} in this example are disjoint sets, but that is not necessary.The number of ways to choose a member of {A, B, C}, and then to do so again, in effect choosing an ordered pair each of whose components are in {A, B, C}, is 3 × 3 = 9.
Originally, a product was and is still the result of the multiplication of two or more numbers.For example, 15 is the product of 3 and 5.The fundamental theorem of arithmetic states that every composite number is a product of prime numbers, that is unique up to the order of the factors.
For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 2 3, a two with a superscript three. In this example, the number two is the base, and three is the exponent. [26] In general, the exponent (or superscript) indicates how many times the base appears in the expression, so that the ...
One says also that a divides b. If a and b are not integers, mathematicians prefer generally to use integer multiple instead of multiple , for clarification. In fact, multiple is used for other kinds of product; for example, a polynomial p is a multiple of another polynomial q if there exists third polynomial r such that p = qr .
The enumerations of Theorems one and two can also be found using generating functions involving simple rational expressions. The two cases are very similar; we will look at the case when , that is, Theorem two first. There is only one configuration for a single bin and any given number of objects (because the objects are not distinguished).
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The theorem says two things about this example: first, that 1200 can be represented as a product of primes, and second, that no matter how this is done, there will always be exactly four 2s, one 3, two 5s, and no other primes in the product.