IB HL Mathematics Homework Counting, Binomial Theorem Solutions
1) How many permutations are there of the letters MATHEMATICS? How many of these permutations begin and end with the letter A? How many of these arrangements do NOT have two vowels adjacent to one another?
Using the permutation formula, we get 11!/(2!2!2!), since there are 2 M's, 2 A's and 2 T's in the word.
Fixing an A in both the beginning and end location, we have 9 other letters to freely permute. This can be done in 9!/(2!2!) ways, accounting for the 2 M's and 2 T's.
There are two ways to interpret the last question – we could count all of the permutations of MATHEMATICS with no adjacent vowels, OR, the just the ones that start and end with A that have no adjacent values. Here are solutions to both:
From all permutations:
There are 7 consonants and 4 vowels. Place the consonants like so:
__ C __ C __ C __ C __ C __ C __ C __
The consonants can be placed in those 7 locations in 7!/(2!2!) ways.
Now, consider placing the vowels in the underlined gaps(__). The two A's can be placed in 8C2 locations, since we will choose 2 out of the 8 locations for the As. Then the E can be placed in any of the 6 remaining locations and the I can be placed in any of the 5 remaining locations.
Using the multiplication principle, we get the answer to be .
From permutations that start and end with A:
There are 7 consonants and 2 vowels left to place. Place the consonants like so:
A C __ C __ C __ C __ C __ C __ C A
Once again, the consonants canb e placed in those 7 locations in 7!/(2!2!) ways.
The E can be placed in 6 places while the I can be placed in the remaining 5 places. The logic here is exactly the same as above, but there is exactly one location for the pair of As. So, the final answer here is 1058400/28 = 37800.
2) A bridge team of four is chosen from six married couples to represent a club at a match. If a husband and wife cannot both be on the team, in how many ways can the team be formed?
There are 12 choices for the first person, followed by 10 for the second (since the first person's spouse is not allowed), followed by 8 for the third and 6 for the fourth. But, we have counted all 4! orders of selecting teams, so we must divide the product 12x10x8x6 by 4! to get our final answer of 12(10)(8)(6)/4! = 240.
Another way to solve this problem is as follows. Choose four of the six couples from which to select the team. This can be done in 6C4 ways. Once we've selected the couples, we have 2 choices for each player from a couple. Thus, there are 24 ways to choose the team, once the four couples are designated. Thus, the total number of teams is (6C4)24 = 240.
3) Jeff is in the portable labyrinth. Luckily, the portables are set up in a grid system. Jeff is currently at grid square (0,0) while his long lost love (Angela, the TOK intern from last semester) is at grid square (8,10). Jeff obviously wants to get to his long lost love as soon as possible, so he will only move in the positive x direction and positive y direction. However, Jeff has neglected to do his Calc BC homework and rumor has it that Worcester will “put the beat down on him” if he sees Jeff. Worcester’s portable is located at grid location (3, 7). So, Jeff would like to avoid crossing through that particular location so that he doesn’t have a black eye when he finally gets to his long lost love. How many ways can Jeff get to his long lost love without a black eye?
The total number of ways in which Jeff can get to Angela is , utilizing the example shown in class that proves that this problem is about choosing 8 of the 18 possible "grid choices" to move in the positive x direction.
Now, we will calculate how many of these paths cross Worcester. There are paths from the current square to Worcester, and there are paths from Worcester to Angela. We can combine any of the first paths with any of the second paths to create a path from Jeff's original location to Angela via Worcester. Thus, the product of these two: is the desired number of paths.
To get the final answer, we much subtract this from the total:
= 37,038.
4)
a) How many four digit numbers do NOT contain any repeating digits? (Note: All four digits numbers are in between 1000 and 9999, inclusive.)
There are 9 choices for the first digi (can't choose 0)t, followed by 9 choices for the second digit (can't choose the first digit), followed by 8 choices for the third digit and 7 choices for the last digit to yield 9x9x8x7 = 4635 such four digit numbers.
b) A number is defined as ascending if each of its digits are in increasing numerical order. For example, 256 and 1278 are ascending numbers, but 1344 and 2687 are not. How many four digit numbers are ascending?
For each combination of four digits out of nine, there is EXACTLY one corresponding ascending number. Thus, the total number of ascending numbers is, the number of ways to choose 4 digits out of 9 (1 through 9).
c) A number is defined as descending if each of its digits are in decreasing numerical order. For example, 652 and 8721 are descending numbers, but 4431 and 7862 are not. How many four digit numbers are descending?
For each combination of four digits out of ten (0 is now allowed), there is EXACTLY one corresponding descending number.Thus, the total number of descending numbers is, the number of ways to choose 4 digits out of 10 (0 through 9).
5) Find the term independent of x in the binomial expansion of .
An arbitrary term in the expansion is .
The value of k that produces the constant term is k=3. We obtain that by setting the exponent to x, 3k-9, to 0. Now, we can obtain the constant term in the expansion:
6) Find the coefficient of x7 in the expansion of (2 + 3x)10, giving your answer as a whole number.
This term is , so the desired coefficient is 2099520.
7) Given that (1+x)5(1 + ax)6 = 1 + bx + 10x2 +…+a6x11, find the values of a and b.
The x term of the expansion is 5x +6 ax = (5+6a)x, which we know is equal to bx.
Thus, b = 5+6a.
The x2 coefficient of the expansion is , which is also 10.
Setting this to 10, we can solve for a:
, so a=0 or a=-2.
Thus, we have two ordered pairs of solutions (a,b): (0,5) and (-2, -7).
8) Find the number of ways in which twelve children can be divided into two groups of six if two particular boys must be in different groups.
Place the two boys in two separate groups, A and B. We are then free to select any five of the children for group A. This can be done in ways. Once this selection is made, the other group is fixed. Thus, there are ways in which the two groups can be subdivided.
9) Students A, B, C, D, E, F, G, H, I, and J must sit in ten chairs lined up in a row. Answer the following questions based on the restrictions given below. (Note that each part is independent of the others, thus no restriction given in part a appliesto the rest of the parts, etc.)
a) How many ways can the students sit if the two students on the ends of the row have to be vowel-named students?
There are three choices for the student on the left, and then 2 choices for the student on the right. Following those two choices, we can arrange the rest of the 8 students left in 8! ways. Thus, the total number of ways the students can sit is (3)(2)(8!).
b) How many ways can the students sit if no two students with vowel names can sit adjacent to each other?
Place all seven consonants like so (C designates an arbitrary consonant):
___ C ___ C ___ C ___ C ___ C ___ C ___ C ___
Now, the empty slots ( ___ ) represent possible locations for the vowels. There are P(8,3) = (8)(7)(6) ways to place the vowels. The 7 consonants can be ordered in 7! ways. Thus, there are (8)(7)(6)(7!) ways the students can sit without any vowel-named students sitting next to each other.
c) Given that students A, B, C, and D are male, and that the rest of the students are female, how many ways can the students be arranged such that the average number of females adjacent to each male is 0.25? (Note: to determine the average number of females each male is adjacent to, sum up the total number of females adjacent to each male and then divide by the total number of males. For example, in the arrangement AEBFCDGHIJ, each male is adjacent to 1.25 females, on average.)
Notice that the only ways in which the average number of females adjacent to males is 0.25 is when all four males are at the left or right end of the row of chairs. If this isn't the case, then more than one female will be adjacent to a male. If this occurs, then the average will be at least 0.5. Since the males and female can sit in any arrangement amongst themselves, for both cases, they can sit in (4!)(6!) ways. Totaling both possibilities (males to the left, males to the right), the total number of arrangements desired is (2)(4!)(6!).
10) There are 7 types of candy sold at Walmart. You are asked to buy 20 bags of candy at Walmart with the following restrictions: you have to get at least three bags of Reece's pieces, you can not get more than four bags of Snickers, and you must get in between three and five bags of Hershey's Kisses, inclusive. How many combinations of bags of candy can you buy?
Go ahead and buy the three bags of Reece's so now, you have 17 bags of candy to buy.
Split the counting into three groups:
1) 3 bags of Hershey's Kisses
2) 4 bags of Hershey's Kisses
3) 5 bags of Hershey's Kisses
For #1, we now have 14 bags left to buy out of six types of candy (since no more Hershey's Kisses are allowed), which can be done in ways. But, of these, we don't want to count the ones that have 5 or more bags of Snickers. Let's count the number of ways to buy 14 bags of six types of candy with 5 or more being Snickers. Go ahead and buy the five Snickers bags, leaving 9 more to buy. This can be done in , thus, the total number of combinations in group 1 is .
For #2, repeat all the logic above, but we start with only thirteen bags left to buy, leading to an answer of .
Finally for #3, repeating the logic again, but with only twelve bags left to buy, we get an answer of .
Adding, we get our final answer to be .
11) Mr. Blue, Mr. Black, Mr. Green, Mrs. White, Mrs. Yellow and Mrs. Red sit around a circular table for a meeting. Mr. Black and Mrs. White must not sit together.
Calculate the number of different ways these six people can sit at the table without Mr. Black and Mrs. White sitting together.
Fix Mr. Black's position. Mrs. White can then sit in three locations, (since she can't sit in Mr. Black's seat, or the two adjacent to him). Then Mr. Blue can pick one of four seats, Mr. Green can pick one of three, Mrs. White can pick one of two, and finally Mrs. Yellow is fixed. Thus, there are a total of 3x4x3x2x1 = 72 ways in which they can sit.