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GRE Subject Test: Math : Combinations

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Example Questions

Example Question #571 : Gre Subject Test: Math

Find C(8,5) .

Possible Answers:

Correct answer:

Explanation:

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C (8,5) = P \frac{(8,5)}{5!}$

$\frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1} =$

The $5 \times 4$ will cancel out because it is in the numerator and denominator,

$\frac{8\times 7\times 6}{3\times 2\times 1} = \frac{336}{6} = 56$ .

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Example Question #11 : Combinations

Find C (10,4) .

Possible Answers:

10,000

210

240

Correct answer:

210

Explanation:

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C (10,4) = P \frac{(10,4)}{4!}$

$\frac{10\times 9\times 8\times 7}{ 4\times 3\times 2\times 1} =$

$\frac{90\times 56}{12\times 2} = \frac{5040}{24} = 210$

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Example Question #81 : Probability & Statistics

Six points are located on a circle. How many line segments can be drawn?

Possible Answers:

Correct answer:

Explanation:

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C=\frac{P(6,2)}{2!}$

There are 2 points on each line segment.

$\frac{6\times 5}{ 2\times 1} = \frac{30}{2} = 15$

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Example Question #582 : Gre Subject Test: Math

Evaluate $_{5}C_{3}$ .

Possible Answers:

Correct answer:

Explanation:

Evaluate $_{5}C_{3}$ is asking to calculate the combination of five objects when choosing three of them.

$_{5}C_{3} = \frac{5!}{(5-3)! 3!}$

$_{5}C_{3} = \frac{5!}{(5-3)! 3!} = \frac{5\times 4\times 3\times 2\times 1}{2\times 1\times3 \times 2\times 1}$

or $3\times2\times1$ cancels out.

$_{5}C_{3} = \frac{5\times 4 }{2\times 1}$

$\frac{20}{2} = 10$

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Example Question #584 : Gre Subject Test: Math

How many ways can I get non-repetitive three-digit numbers from the numbers: $\{2,3,4,5,6,7,8,9,0\}$ ?

Possible Answers:

443

448

451

441

Correct answer:

448

Explanation:

Step 1: Count how many numbers I can use..

I can use 9 numbers.

Step 2: Determine how many numbers I can put in the first digit of the three-digit number..

I can put  numbers in the first spot. I cannot put  in the first slot because the number will not be a three-digit number.

Step 3: Determine how many numbers I can put in the second digit..

I can also put  numbers in the second spot. Here's the reason why it's still :

Let's say I choose 2 for the first number. I will take out of my set. I had numbers in my set..If i take a number out, I still have  numbers left. These numbers are: 3,4,5,6,7,8,9,0 .

Step 4: Determine how many numbers I can put in the third and final digit...

I can put  numbers in the third slot..

I had  numbers at the start, and then I removed  of them. 9-2=7 .

Step 5: Multiply how many numbers can go in the first, second, and third spot..

$8\cdot8\cdot7=448$ .

There are a total of 448 non-repetitive three-digit numbers that can be formed.

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Example Question #13 : Combinations

A coach of a baseball team needs to choose players out of a total players in the team. How many ways can the coach choose 9 players?

Possible Answers:

24,310

21,430

21,340

23,410

Correct answer:

24,310

Explanation:

Step 1: Recall the combination formula...

$_nC_r=\frac {n!}{r!(n-r)!}$

Step 2: Find and from the question..

n=17, r=9 .

Step 3: Plug in the values into the formula above..

$\frac {17!}{9!(17-9)!}=\frac {17!}{9!\times8!}=24,310$

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Example Question #82 : Probability & Statistics

Six people run in a race, in how many different orders can they finish?

Possible Answers:

300

46,656

720

Correct answer:

720

Explanation:

This problem is solved by knowing that we have six options for first place, five options for second place, and so on.

Which means

6*5*4*3*2*1 = 720

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GRE Subject Test: Math : Combinations

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #571 : Gre Subject Test: Math

Example Question #11 : Combinations

Example Question #81 : Probability & Statistics

Example Question #582 : Gre Subject Test: Math

Example Question #584 : Gre Subject Test: Math

Example Question #13 : Combinations

Example Question #82 : Probability & Statistics

All GRE Subject Test: Math Resources

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