Cracking the SAT subject test in math 2, 2nd edition

Cracking the SAT Subject Test in Math 2, 2nd Edition one basket in three attempts? (A) (B) (C) (D) 1 (E) 45 A jar contains b blue marbles, r red marbles, and g green marbles If two marbles are to be t[.]

(A)

(B)

(C) (D) 1 (E)

45 A jar contains b blue marbles, r red marbles, and g green

marbles If two marbles are to be taken from the jar one

at a time, then what is the probability that the first marble is blue and the second marble is red?

(A)

(B)

(C)

(D)

(E)

PERMUTATIONS, COMBINATIONS, AND

FACTORIALS

Questions about permutations, combinations, and factorials are relatively

rare on the SAT Subject Test in Math 2 As with many of the topics from precalculus, most of these questions are not as hard as they look Rather, these questions test your knowledge of these topics and your ability to work with them Both permutations and combinations are simply ways of counting groups, whereas factorials are a mathematical operation which arises from permutations and combinations (though factorials may be tested separately from permutations and combinations on the SAT Subject Test in Math 2

Simple Permutations

A permutation is an arrangement of objects of a definite order The simplest sort of permutation question might ask you how many different arrangements are possible for 6 different chairs in a row, or how many different 4-letter arrangements of the letters in the word FUEL are possible Both of these simple questions can be answered with the same technique

Just draw a row of boxes corresponding to the positions you have to fill

In the case of the chairs, there are six positions, one for each chair You would make a sketch like the following:

Then, in each box, write the number of objects available to put into that box Keep in mind that objects put into previous boxes are no longer available For the chair-arranging example, there would be 6 chairs available for the first box; only 5 left for the second box; 4 for the third, and so on until only one chair remained to be put into the last position Finally, just multiply the numbers in the boxes together, and the product will be the number of possible arrangements, or permutations

There are 720 possible permutations of a group of 6 chairs This number can also be written as “6!”—that’s not a display of enthusiasm—the

exclamation point means factorial The number is read “six factorial,”

and it means 6 • 5 • 4 • 3 • 2 • 1, which equals 720 A factorial is simply the product of a series of integers counting down to 1 from the specified number For example, the number 70! means 70 • 69 • 68…3 • 2 • 1

The number of possible arrangements of any group with n members is simply n! In this way, the number of possible arrangements of the letters

in FUEL is 4!, because there are 4 letters in the group That means 4 • 3 •

2 • 1 arrangements, or 24 If you sketched 4 boxes for the 4 letter positions and filled in the appropriate numbers, that’s exactly what you’d get

Advanced Permutations

Permutations get a little trickier when you work with smaller arrangements For example, what if you were asked how many 2-letter arrangements could be made from the letters in FUEL? It’s just a modification of the original counting procedure Sketch 2 boxes for the 2 positions Then fill in the number of letters available for each position As before, there are 4 letters available for the first space, and 3 for the second The only difference is that you’re done after two spaces

As you did before, multiply the numbers in the boxes together to get the total number of arrangements You should find that there are 12 possible 2-letter arrangements from the letters in FUEL

That’s all there is to permutations The box-counting procedure is the safest way to approach them Just sketch the number of positions available, and fill in the number of objects available for each position, from first to last—then multiply those numbers together

On to Combinations

Combinations differ from permutations in just one way In combinations,

order doesn’t matter A permutation question might ask you to form different numbers from a set of digits Order would certainly matter in that case, because 135 is very different from 513 Similarly, a question about seating arrangements would be a permutation question, because the word “arrangements” tells you that order is important So questions that ask about “schedules” or “orderings” require you to calculate the

number of permutations.

Combination questions, on the other hand, deal with groupings in which order isn’t important Combination questions often deal with the selection of committees Josh, Lisa, Andy isn’t any different from Andy, Lisa, Josh, as far as committees go In the same way, a question about the number of different 3-topping pizzas you could make from a 10-topping list would be a combination question, because the order in which the toppings are put on is irrelevant Questions that refer to “teams” or

“pairs” are therefore asking about the number of possible combinations.

Which One to Use?

Combination and permutation questions can be very similar in appearance Always ask yourself carefully whether sequence is important in a

certain question before you proceed.

Calculating Combinations

Calculating combinations is surprisingly easy All you have to do is throw out duplicate answers that count as separate permutations, but not as separate combinations For example, let’s make a full-fledged combination question out of that pizza example

pepperoni meatballs

green peppers mushrooms tomato sausage anchovies onion garlic broccoli

21 If a pizza must have 3 toppings chosen from the list

above, and no topping may be used more than once on a given pizza, how many different kinds of pizza can be made?

(A) 720 (B) 360 (C) 120 (D) 90 (E) 30

Here’s How to Crack It

To calculate the number of possible combinations, start by figuring out

the number of possible permutations.

That tells you that there are 720 possible 3-topping permutations that can be made from a list of 10 toppings You’re not done yet, though Because this is a list of permutations, it contains many arrangements that duplicate the same group of elements in different orders For example, those 720 permutations would include these:

pepperoni, mushrooms, onion pepperoni, onion, mushrooms mushrooms, pepperoni, onion mushrooms, onion, pepperoni onion, pepperoni, mushrooms

All six of these listings are different permutations of the same group In fact, for every 3-topping combination, there will be 6 different permutations You’ve got to divide 720 by 6 to get the true number of combinations, which is 120 The correct answer is (C)

So, how do you know what number to divide permutations by to get combinations? It’s simple For the 3-position question above, we divided

by 6, which is 3! That’s all there is to it To calculate a number of possible combinations, calculate the possible permutations first, and divide that number by the number of positions, factorial Take a look at one more:

14 How many different 4-person teams can be made from a

roster of 9 players?

(A) 3,024 (B) 1,512 (C) 378 (D) 254 (E) 126

Here’s How to Crack It

This is definitely a combination question Start by sketching 4 boxes for the 4 team positions

Then fill in the number of possible contestants for each position, and multiply them together This gives you the number of possible

permutations.

Finally, divide this number by 4! for the 4 positions you’re working with This gets rid of different permutations of identical groups You divide 3,024 by 24 and get the number of possible combinations, 126 The correct answer is (E)

Using Your Calculator to Solve Permutations

and Combinations

On the TI-84, you can quickly solve permutations and combinations using “nPr” (for permutations) and “nCr” (for combinations), found under the PRB submenu of the MATH menu

To use these tools, first you need to determine the “n” and “r” values For both, n is the total number of options you are choosing from, and r is the number of choices you are making First, input your n value Then, press

MATH->PRB and select either the second (for permutations) or third (for

combinations) option Then enter your r value.

For example, let’s say you have 15 dragon figurines and you want to display 7 of them in a row To determine the number of ways you can make your display, use nPr (because order matters) and input the following into your calculator:

15 nPr 7 = 32,432,400

Now, if you wanted to find the number of groups of 7 dragon figurines you could make from your collection, you would use nCr instead:

15 nCr 7 = 6,435

In both cases, the approach of making slots and filling in the options will work However, knowing how to use these calculator shortcuts can save you time, especially when you are dealing with bigger numbers on the SAT Subject Test in Math 2

On the SAT Subject Test in Math 2, ETS occasionally asks you to calculate

a factorial itself If you try to do a factorial question in your head, you’re likely to fall into one of ETS’s traps Use your calculator and be careful

(A) !

(B) (C) (D) 5 (E)

For this question, just use your calculator 5! = 120 and 6! = 720, so you have The answer is (C) It’s supposed to be easy, but don’t try to simplify this in your head In this instance, every option is a potential trap, so take the time to be sure about your choice

Factoring the Factorial

Sometimes numbers will be too bulky for your calculator, or you’ll realize there’s a faster way You can factor factorials Let’s take another look at question 18 from the previous page Notice that the denominator is 6! − 5! 6! is the same as 6 • 5!, which means you can

factor 5! out of the denominator and you’re left with 5!(6 − 1) Now you can cancel and you end up with

Let’s try a harder problem:

(A)

(B) n(n − 1)!

(C) (n!)2

(D)

(E) (n−1)!+n

Here’s How to Crack It

As you’ve seen throughout this book, Plugging In is a great way to make

your life much easier Make n = 5 Now the problem is

Now you can multiply and you find that Plug 5 into each answer choice for n; only (D)

equals 4

AND FACTORIALS

Try the following practice questions about permutations, combinations, and factorials The answers can be found in Part IV

12 How many different 4-student committees can be

chosen from a panel of 12 students?

(A) 236 (B) 495 (C) 1,980 (D) 11,880 (E) 20,736

18 In how many different orders may 6 books be placed on

a shelf?

(A) 36 (B) 216 (C) 480 (D) 720 (E) 46,656

31 How many 7-person committees consisting of 4 females

and 3 males may be assembled from a pool of 17 females and 12 males?

(A) 523,600 (B) 1,560,780 (C) 1.26 × 107 (D) 7.54 × 107 (E) 7.87 × 109

49 If there are n available substitute teachers at a school

and n > 3, then how many groups of 3 substitute