How Many Ways Can a Group of 8 Students Be Selected From 9 Students?

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From a group of viii students, 3 are attention a meeting.

Quantity A: The number of different groups that could attend among the viii students

Quantity B: 336

Possible Answers:

Quantity A is greater.

The relationship cannot be determined from the information given.

Quantity B is greater.

The two quantities are equal.

Correct reply:

Quantity B is greater.

Explanation:

To solve this problem, you would need to utilize the combination formula, which is $\dpi{100} \small C=\frac{n!}{r!(n-r!)}$ .

$\dpi{100} \small C$ is the number of possibilities, $\dpi{100} \small n$ is the number of students, and $\dpi{100} \small r$ is the students attending the meeting. Thus, $\dpi{100} \small \frac{8!}{3!5!}=56$ .

336 would be the result of calculating the permutation, which would exist incorrect in this instance.

Michael owns 10 paintings. Michael would like to hang a single painting in each of v dissimilar rooms. How many unlike means are there for Michael to hang five of his 10 paintings?

Correct answer:

$\dpi{100} \small 30,240$

Explanation:

This problem involves the permutation of 10 items beyond 5 slots. The first slot (room) tin have x different paintings, the second slot can have 9 (one is already in the commencement room), the third slot can have viii and so on. The number of possibilities is obtained by multiplying the number of possible options in each of the 5 slots together, which here is $\dpi{100} \small 10\times 9\times 8\times 7\times 6=30,240$ .

How many different committees of 3 people tin be formed from a group of vii people?

Correct answer:

Explanation:

There are $\dpi{100} \small 7\times 6\times 5=210$ different permutations of 3 people from a grouping of seven (when gild matters). There are $\dpi{100} \small 3\times 2=6$ possible ways to conform iii people. Thus when club doesn't matter, there tin can be $\dpi{100} \small \frac{210}{6}=35$ dissimilar committees formed.

A restaurant has a meal special that allows you lot to choose one of three salads, 1 of five sandwiches, and 2 of fifteen side dishes. How many possible combinations are there for the repast?

Correct respond:

1575

Explanation:

Although this is a permutation way problem, we have to be careful regarding the last portion (i.e. the side dishes). We know that our meal volition have:

(3 possible salads) * (5 possible sandwiches) * (x possible combinations of side dishes).

We must ascertain how many side dishes can be selected. At present, it does not matter what gild we put together the side dishes, so we have to apply the combinations formula:

c(north,r) = (n!) / ((n-r)! * (r!))

Plugging in, we get: c(15,2) = 15! / (13! * 2!) = 15 * 14 / 2 = 15 * 7 = 105

Using this in the equation in a higher place, we get: iii * five * 105 = 1575

The President has to choose from vi members of congress to serve on a committee of three possible members. How many different groups of three could he choose?

Explanation:

This is a combinations problem which means guild does not matter. For his first choice the president can choose from 6, the second 5, and the 3rd 4 so y'all may remember the answer is 6 * v * iv, or 120; yet this would be the answer if he were choosing an ordered set similar vice president, secretary of country, and primary of staff. In this case lodge does not matter, so you must divide the vi * five * 4 by three * 2 * 1, for the iii seats he's choosing. The answer is 120/six, or 20.

Mohammed is being treated to ice cream for his birthday, and he'south allowed to build a three-scoop sundae from any of the thirty-one available flavors, with the only condition being that each of these flavors be unique. He's also immune to pick different toppings of the available, although he's already decided well in advance that i of them is going to be peanut butter cup pieces.

Knowing these details, how many sundae combinations are available?

Correct respond:

40455

Caption:

Considering order is not of import in this problem (i.e. chocolate flake, pecan, butterscotch is no different than pecan, butterscotch, chocolate chip), it is dealing with combinations rather than permutations.

The formula for a combination is given as:

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

where is the number of options and is the size of the combination.

For the ice cream choices, at that place are thirty-one options to build a three-scoop sundae. So, the number of ice cream combinations is given every bit:

$C(31,3)=\frac{31!}{(31-3)!3!}=4495$

Now, for the topping combinations, we are told there are 10 options and that Mohammed is allowed to option two items; however, nosotros are too told that Mohammed has already chosen one, so this leaves nine options with one item existence selected:

$C(9,1)=\frac{9!}{(9-1)!1!}=9$

And then there are nine "combinations" (using the word a fleck loosely) available for the toppings. This is perhaps intuitive, but it'south worth doing the math.

At present, to discover the total sundae combinations—ice cream and toppings both—we multiply these ii totals:

4495(9)=40455

If in that location are students in a form and people are randomly choosen to go class representatives, how many different ways can the representatives exist chosen?

Correct answer:

190

Explanation:

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the ii course representatives are randomly chosen, therefore information technology doesn't matter what order the representative is chosen in, the end result is the same. The general formula for combinations is $C(n,k)=\frac{n!}{(n-k)!k!}$ , where is the number of things yous take and is the things yous desire to combine.

Plugging in choosing 2 people from a group of 20, we find

$C(20,2)=\frac{20!}{(20-2)!2!}=190$ . Therefore there are a 190 different means to choose the class representatives.

There are eight possible flavors of curry at a particular eating place.

Quantity A: Number of possible combinations if 4 unique curries are selected.

Quantity B: Number of possible combinations if five unique curries are selected.

Possible Answers:

The 2 quantities are equal

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

Correct respond:

Quantity A is greater.

Explanation:

The number of potential combinations for selections made from possible options is

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

Quantity A:

$C(8,4)=\frac{8!}{(8-4)!4!}=70$

Quantity B:

$C(8,5)=\frac{8!}{(8-5)!5!}=56$

Quantity A is greater.

Quantity A: The number of possible combinations if 4 unique choices are made from ten possible options.

Quantity B: The number of possible permutations if two unique choices are fabricated from x possible options.

Possible Answers:

The relationship cannot be established.

Quantity A is greater.

The two quantities are equal.

Quantity B is greater.

Correct reply:

Quantity A is greater.

Explanation:

For choices made from possible options, the number of potential combinations (order does non affair) is

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

And the number of potential permutations (order matters) is

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

Quantity A:

$C(10,4)=\frac{10!}{(10-4)!4!}=210$

Quantity B:

$P(10,2)=\frac{10!}{(10-2)!}=90$

Quantity A is greater.

In that location are possible flavor options at an ice cream shop.

$\textup{Quantity A: The number of possible combinations if three unique flavors are selected.}$

$\textup{Quantity B: The number of possible combinations if five unique flavors are selected.}$

Correct answer:

$\textup{Quantity B is greater}$

Caption:

When dealing with combinations, the number of possible combinations when selecting choices out of options is:

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

For Quantity A, the number of combinations is:

$C(10,3)=\frac{10!}{(10-3)!3!}$

C(10,3)=120

For Quantity B, the number of combinations is:

$C(10,5)=\frac{10!}{(10-5)!5!}$

C(10,5)=252

Quantity B is greater.

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How Many Ways Can a Group of 8 Students Be Selected From 9 Students?

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All GRE Math Resources

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