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- Mike is arranging seven of his various awards and commendations on a shelf in his office. If he insists that his hard-fought Class of 1999 Math Award be placed in the center, in how many different orders could he arrange the seven items?
(A) 60
(B) 72
(C) 120
(D) 720
(E) 1440
The best way to tackle a possibilities problem like this is to draw a bunch of blanks like you're about to play Hangman, and then start thinking, methodically, through the choices you have at every step along the way. I'm going to illustrate this process with slightly more thoroughness than you probably will on test day (you won't need to make up award names, but I will because it's hilarious):
Position
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1
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2
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3
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4
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5
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6
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7
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Award
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Class of 1999 Math Award
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Choices
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1
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First, as I did above, you must account for any special conditions or restrictions. The Math Award must go in the middle, so there's only one choice for Position 4.
Position
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1
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2
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3
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4
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5
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6
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7
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Award
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Invisible Man Award
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Class of 1999 Math Award
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Choices
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6
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1
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Once all the restrictions are accounted for, start filling in the rest of the spaces. To fill Position 1, Mike has 6 different awards to choose from. Say, for argument's sake, he chooses the Invisible Man Award next.
Position
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1
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2
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3
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4
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5
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6
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7
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Award
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Invisible Man Award
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Acne League – Most Improved
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Push Ups Contest – Last Place
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Class of 1999 Math Award
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Spin the Bottle – Luckiest Player Ever
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5th Grade Science Fair – 2nd Place
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Shortest Fight in School History – Loser
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Choices
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6
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5
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4
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1
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3
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2
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1
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To fill the next spot, since he used up the Invisible Man Award, he has 5 awards left to choose from. Once he chooses the Acne League – Most Improved award for Position 2, he has 4 choices for Position 3. He continues this process until he's filled all the positions.
To calculate the number of different arrangements Mike could have made, multiply the number of choices he had at every step:
6 × 5 × 4 × 1 × 3 × 2 × 1 = 720
When solving a possibilities problem, set up the hangman blanks, then imagine yourself actually performing the task described. First take care of special conditions or restrictions, and then take care of everything else. At every step, ask yourself “How many choices do I have?” And then ask yourself “How many choices do I have now?” And then – you guessed it – ask yourself how many choices you have again.
Stop when you run out of choices.
Possibly you're interested in more practice?
- Alicia is arranging photographs of five family members in a row on her refrigerator. If she wants the photograph of her mother and the photograph of her father to be on opposite ends of the row, how many arrangements for the photographs are possible?
(A) 120
(B) 60
(C) 24
(D) 21
(E) 12
- Brady is about to pull four cards at random from a deck of ten unique cards, and line them up on a table. What is the number of possible card arrangements that Brady could make?
(A) 24
(B) 240
(C) 1456
(D) 5040
(E) 6220
Answers:
17. E
19. D
*Aside from this footnote, I'm deliberately avoiding the terms “permutation,” “combination,” and “factorial” here. That's not because I don't know them; it's because I've found that they manufacture more confusion than they alleviate on the SAT. Remember: The SAT is not a math test! If you prefer to solve the questions laid out here by cramming them into nPr and nCr notations in your calculator, be my guest, but don't cry to me when you miss counting questions on the SAT.
Do you think you could explain number 19 a bit? I'm still confused on that one.
ReplyDeleteSure! The basic idea is that there are 10 cards that could be pulled for when he chooses the first card, then 9 choices left for the second pull, 8 choices left for the 3rd pull, and 7 choices left for the fourth.
ReplyDelete10*9*8*7 = 5040.
Remember, the important question to ask yourself at every choice is: "How many choices do I have?"
Can't you just take the factorial of 6? I remember a long time ago my Algebra teacher told me that possibility questions could be solved with factorials.
ReplyDeleteYes, the awards problem simplifies to 6!. But that's only because there's only one constraint on it. It would get more complicated if, for example, 3 of the awards needed to go in the middle 3 spots, and the other awards could go anywhere. You could still reduce the solution to factorials (3!4!), but i doubt it would be any faster than just listing out the possibilities for each position as done above.
ReplyDeleteFactorials, and combinatorics in general, become more necessary when problems involve bigger numbers. On the SAT, the numbers will always remain small, so I choose not to use a chainsaw when a butter knife will do. :)
for number 16, isn't it suppose be like 1 3 2 1 1 ?
ReplyDeleteSorry, which question are you referring to? There's no #16 in this post.
ReplyDeleteFor number 17. I don't understand it. Let us have ABCDE, for the five family members respectively. A and F have only 1 possibility. Then, for BCD in the middle, there are 3*2*1 possibilities. Why is the answer 12?
ReplyDeleteAlicia's mom and dad just have to be on opposite ends. So Alicia could put her mom first and her dad last, or she could put her dad first and her mom last.
ReplyDeleteSo it's really like this:
2 choices for first position, then 1 choice for last position.
Then you fill in the middle positions (as you did correctly): 3 choices, 2 choices, one choice.
2 × 3 × 2 × 1 × 1 = 12
Oh. I thought it was her mom first and then dad last. I didn't think about the second option. Do you have any more questions with probability? Also, I am a 7th grader and I just took my SAT on Jan 26. I received a 660 in math and a 1910 overall. However, I think a LOT of my mistakes in math are careless. I just did your diagnostic 1 and I got 17/20. 2 of them were careless; one was not.
ReplyDeleteI have a post about probability, and I think there are a few more in the diagnostic drills, but I'd caution you against worrying TOO much about probability (for the SAT, at least). Probability questions typically only occur about once per test, and HARD probability questions are very rare.
ReplyDelete