Permutation Problems, 2

Category: Algebra, Statistics

"Published in Suisun City, California, USA"

A man bought three vanilla ice cream cones, two chocolate cones, four strawberry cones, and five butterscotch cones for his 14 children. In how many ways can he distribute  the cones among his children?

Solution:

The given word problem above is about permutations but it is a different type which is called a Distinguishable Permutation.  

If a set of n objects consists of k different kinds of objects with n₁, objects of the first kind, n₂ objects of the second kind, n₃ objects of the third kind, and so on, where n₁ + n₂ + ......... + n_k = n, then the number of distinguishable permutations of these objects is 

Now, let's go back to the given problem, if n = 14 children, n₁ = 3 vanilla ice cream cones, n₂ = 2 chocolate cones, n3 = 4 strawberry cones, and n₄ = 5 butterscotch cones, then the number of ways to distribute the cones among to his children is

  

Permutation Problems

Category: Algebra, Statistics

"Published in Newark, California, USA"

If polygons are labeled by placing letters at their vertices, how many ways are there of labeling (a) a triangle, (b) a quadrilateral, (c) a hexagon with the first 10 letters of the alphabet?

Solution:

The given word problem above is about permutations. Permutation is an arrangement of a number of objects in a definite order. To "permute" a set of objects means to arrange them in a definite order. The number of permutations of n things taken r at a time is given by the formula

where n! (read as n factorial) is equal to n(n -1)(n - 2)......3∙2∙1. Take note that the values of n and r must be zero and positive numbers only. 0! is equal to 1. 

Now, let's go back to the given problem and solve for the permutations of the given polynomials. 

(a) For a triangle, the number of ways to label the vertices with the first 10  letters of the alphabet are

(b) For a quadrilateral, the number of ways to label the vertices with the first 10  letters of the alphabet are

 (c) For a hexagon, the number of ways to label the vertices with the first 10  letters of the alphabet are

Binomial Theorem, 4

Category: Algebra

"Published in Newark, California, USA"

Find the term which does not contain x in the expansion for

Solution:

Consider the formula in getting the value of rth term

In this problem, we need only x and y in order to solve for the value of r where x in not involve in the binomial expansion. If x is not involve in the binomial expansion, then the exponent of x is 0. Any number (except zero) raised to zero power is always equal to one.

Since we need the exponents of x in order to solve for the value of r, then we can omit y and 2 as follows

Take the logarithm on both sides of the equation to the base x, we have

Therefore, there's no x at the 3rd term.Consider again the formula in getting the value of rth term

Substitute the values of n, r, x, and y in order to get the value at the 3rd term, we have