Combinations/Permutations CHALLENGE! Win Free Subscription!

[chrismagoosh]

Here are five difficult combinations/permutations problems.

 

If you can get all 5 right, you will win a free subscription to magoosh GRE!

 

EDIT BY MODERATOR

 

I will be taking answers until Fri.

 

 

 

 

1. A committee of three must be formed from 5 women and 5 men. What is the probability that the committee will be exclusive to one gender?

 

(A) 1/60

(B) 1/120

© 1/8

(D) 1/6

(E) 1/3

 

2. A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D)?

 

(A) 1/55

(B) 1/66

© 2/17

(D) 1/110

(E) 2/55

 

 

3. A homework assignment calls for students to write 5 sentences using a total of 10 vocabulary words. If each sentence must use two words and no words can be used more than once, then how many different ways can a student select the words?

 

(A) 10!/5!

(B) 10!/32

© 5! x 5!

(D) 2! x 5!

(E) 10!

 

4. Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people.

 

Column A

Number of unique team S

 

Column B

Number of unique Team R

 

 

5. A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander’s decision, then which of the following cannot be the value of p?

 

(A) x

(B) x + 3

© 3

(D) 6

(E) 8

Edited March 21, 2011 by wasleys
Removed spammy link

[kamakshi]

1 [D]

2

3 [D]

 

4 and 5 i will answer later. I am running out of time now,

[kamakshi]

4 is [D]

 

I think 5 is 12Cx = (12+P)CX and we got to solve for P

[chrismagoosh]

You got two of them right :) (I'm not sure if you gave an answer for #5.)

 

I appreciate the effort. Have another go at them.

 

P.S. -- I made the problems similar to a Computer Adaptive Test, assuming you are getting them all right. That is, the problems get progressively more difficult with #5 be the most difficult of all.

 

Good luck guys!

[kamakshi]

3 4 and 5 [E]. Please let me know if my answers are correct. I can explain how I solved the problems. By the way, thank you for posting such good questions.

[nkt06]

Not sure if these are right, but here are my answers:

 

1.

= (5C3+5C3) / 10C3

= 20 / 120

= 1/6

=> ANSWER D

 

2.

A-L = 12 letters

20 desirable possibilities [10 for forward 3-letter sequences (like ABC) 10 for backward 3-letter sequences (like FED)]

 

12*11*10 = total possibilities

 

=> 20/(12*11*10) = 1/66

=> ANSWER B

 

3.

= 10P2 * 8P2 * 6P2 * 4P2 * 2P2

= 10*9*8*7*6*5*4*3*2

= 10!

=> ANSWER E

 

4.

= x! / (n!)(x-n)! v (x+1)! / (n+1)! (x-n)!

= 1 v (x+1) / (n+1)

= n+1 v (x+1)

= n v x

=> Answer depends on values of n and x

=> ANSWER D

 

5.

12! / (x!)(12-x)! = 12! / (x+p)!(12-x-p)!

=> x(12-x) = (x+p)(12-x-p), where x and p are integers

When p = 3, we get:

6x = 27 => x = 9/2 which is not an integer, therefore p != 3

=> ANSWER C

[appreciate]

1. D

2. B

3. E

4. B

5. C

[appreciate]

1. 5C3+5C3/10C3

2. 20/12P3 [A-J=10 possibility and reversely another 10 possibilities]

3. 10P2*8P2*6P2* 4P2 * 2P2= 10!

4. XCn X+1Cn+1

n+1 X+1

obviously X+1>n+1, So, Ans is B

5.12Cx+p=12Cx

or, 12=x+p+x=2x+p

putting the value of P in the 1st relation, All the value satisfy the relation except, © P=3

[nkt06]

For #4, theoretically x could be equal to n, no? For example, you could be choosing 5 people from 5 people, so 4 still depends on n and x? Still D?

[kamakshi]

I believe #4 is to do with Permutations and not combinations because question is asking for unique combinations and so I believe that would be the answer.

 

For Q 5 if P astronauts are added why wouldn't the total astronauts as well increase by 12+P?

[chrismagoosh]

There is a lot of great work here! Kamakshi one of your answers from 3-5 is right. nkto6 you got 4 out of 5 right. Appreciate you were close behind with 3.

Thanks as well as providing all the work. I will say this much, if I combine everyone's answers you have the correct answers. That is, no one question

confounded everyone.

[nkt06]

Ahhhh for #3 you said "select" not "arrange". This implies combinations, not permutations.

 

So,

= 10!/(2!8!) * 8!/(2!6!) * 6!/(2!4!) * 4!/(2!2!) * 2!/(2!0!)

= 10! / (2^5)

= 10! / 32

=> ANSWER B

[appreciate]

Please give the correct ans with explanation. We will be sort out our mistake.who will get the free subscription, please?

[chrismagoosh]

nkt06 got it!

 

It was question number 3 that was stumping people the most. So, yes you are selecting pairs of words--their order doesn't matter once they're selected so you get

10C2 x 8C2...2C2 giving you 10!/2^5, or 10!/32. Ans (B).

 

The other toughie was the last one was number 4. In this case if you make n = x and pick the number 3, you get 3C3 = 1. Then n + 1, x + 1, would be 4C4 = 1.

In this case the columns are equal. If x does not equal n so that you have 5C3 vs 6C4 (I'm just picking random numbers) the right side will always be

bigger. You can't have both answers (B) and © so the answer is (D).

 

Good job again everyone!

 

nkt06 could you give me your email via my urch profile? I'll send you a subscription code.

[nkt06]

Thanks chrismagoosh! I just sent you a message.

[twohundredping]

Please give the correct ans with explanation.

 

In this problem ncr(n,r) = n!/(r! * (n-r)!) and npr(n,r) = n!/(n-r)!

 

I don't have any affiliations with this guy but I'm guessing nkt06 will be the only guy given a subscription. Now that someone was awarded the subscription I can give you the answers. By no means are any of my answers the "correct" way to do the problem. Its just the first idea that popped into my head.

 

1) A committee of three must be formed from 5 women and 5 men. What is the probability that the committee will be exclusive to one gender?

 

2 * ncr(5,3) * ncr(5,0) / ncr(10,3) = (2*5*4/2*1) / (10*9*8/6) = 1/6

 

 

2) A three-letter code is formed using the letters A-L, such that no letter is used more than once. What is the probability that the code will have a string of three consecutive letters (e.g. A-B-C, F-E-D) ?

 

So clearly order is important. We must choose in order A,B,C or H,G,F etc.

So there are 12 letters in this set. Lets break this problem into parts.

If the first letter we select is anywhere from C - J then we have two directions to go for the second letter and one direction to go for the final letter = 8 * 2 * 1

If the first letter we select is A,B,K,L then we have one direction to go for the second

letter and one direction to go for the third letter = 4*1*1

Therefore, we have (8 * 2 * 1 + 4 * 1 * 1) / npr(12, 3) = 20 / (12 * 11 * 10) = 1/66

 

 

Ok so lets do it with an order is not important formula (ncr) to make it more difficult.

So, we have the sets (A,B,C), (B,C,D), ... , (J,K,L). There are 10 of these.

Consider the set (A,B,C):

Order not important lets it be (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A)

Notice that only 2/6 = 1/3 of these are valid.

So we have (10 * 1/3) / ncr(12, 3) = (10 / 3) / (12 * 11 * 10 / 6) = 2 / (12 * 11) = 1/66

 

 

3) A homework assignment calls for students to write 5 sentences using a total of 10 vocabulary words. If each sentence must use two words and no words can be used more than once, then how many different ways can a student select the words?

 

Its important to realize the question does not ask how many ways the sentences can be written, just how many ways the student can select the words (order of the two words selected thus will not be important so we use ncr)

 

ncr(10,2) * ncr(8,2) * ncr(6,2) * ncr(4,2) * ncr(2,2) = 10! / (2!)^5 = 113400

 

 

4) Team S is to comprise of x debaters chosen from n people? Team R is comprised of x + 1 debaters chosen from n+1 people.

 

A: Number of unique team S

B: Number of unique Team R

 

Consider n = x, ncr(n,x) = ncr(n+1,x+1) = 1 and thus A = B

If x

So this cannot be determined

 

 

Lets make this problem harder. Add the restriction that x

ncr(n,x) = n * (n-1) * .... * (x+1) / (n-x)!

ncr(n+1,x+1) = (n+1) * n * ..... * (x+2) / (n+1 - (x+1))! = (n+1) * n * .... * (x+2) / (n-x)!

So in this case, clearly column B is greater (I of course assume that n,x >= 0)

 

 

5) A lunar mission is made up of x astronauts and is formed from a total of 12 astronauts. A day before the launch the commander of the program decides to add p astronauts to the mission. If the total number of possible lunar missions remain unchanged after the commander’s decision, then which of the following cannot be the value of p?

 

(A) x

(B) x + 3

© 3

(D) 6

(E) 8

 

Remember that ncr(n,x) = ncr(n,n-x)

Lets try to make these values fit the above statement to prove that they CAN be equal

Can 2x = 12 - x? Yes, x = 4

Can 2x + 3 = 12 - x? Yes, x = 3

Can x + 3 = 12 - x? No, this reduces to 2x = 9. We have to have integer x

Can x + 6 = 12 - x? Yes, x = 3

Can x + 8 = 12 - x? Yes, x = 2

Therefore, the answer is C

 

(You may notice a logical fault with how this problem is worked but this is the GRE and we can always use elimination)

[kamakshi]

Can someone answer this permutation and combination question?

 

A person went to the market with a red and a green bag and bought 10 carrots and 6 radhishes. On the way to home he distributed vegetables into the two bags in such a way that no bag is empty. In how many ways can he do this?

 

is it 10c2 * 6c2

 

or

 

2 power 16 - 2

[twohundredping]

We will do this problem considering that order is not important to make it easier (the way most people try to do these problems). So the order of vegetables in each bag is not important, just the distinct vegetables in each bag. This means we will use n choose k (nck) instead of n pick k (npk). Please note that nck and npk are the same thing as the ncr and npr I posted in post 16 except now I'm using the variable k (more conventional) instead of r. A key part to this problem is to ignore the second bag. It just gets the leftovers and there is only one way to order those leftovers for any given choice of vegetables in the first bag. So we just consider the first bag and let it have anywhere from 1 to 15 vegetables (we don't need to distinguish between carrots and radishes). We have 1 to 15 vegetables in the first bag since neither bag can be empty.

 

The answer will be 16 choose 1 + 16 choose 2 + ... + 16 choose 15 = sum(16 choose k, k, 1, 15)

 

I had to look this next formula up but thats not important. I just did it so I would not have to use a calculator. The GRE would not require you to know this specialized formula and if you had to do this problem it would be alot less than 16 vegetables so that you could calculate each n choose k. I just mention this because I am a fundamentalist and I think that the less formulas I have in my head the better.

 

sum(n choose k, k, 0, n) = 2^n

So we can alter this to sum(n choose k, k, 1, n-1) = 2^n - (n choose 0) - (n choose n) = 2^n - 2

Therefore there are 2^16 - 2 = 65536 - 2 = 65534 (assuming I still know my powers of 2)

 

 

If you are having problems understanding this problem then make it a bit simpler. Let it be 4 vegetables.

Example with 4 items... Equation gives 14

Bag 1    Bag 2

a        bcd
ab       cd
ac       bd
ad       bc
abc      d
abd      c
acd      b

I could keep enumerating all cases but notice the symmetry. Any other combinations of vegetables in bag 1 have already been listed in bag 2. Therefore, we get 7 * 2 = 14. This is not just guesswork on my part. Consider that if I list all the ways vegetable 'a' can be the first bag + all the ways vegetable 'a' can be in the second bag I get the whole set. I'm a bit in a rush now but if you want any more detail don't hesitate to ask.

Edited March 24, 2011 by twohundredping

[databreath]

Somehow, I'm not getting this for question #5::::::

Remember that ncr(n,x) = ncr(n,n-x)

Lets try to make these values fit the above statement to prove that they CAN be equal

Can 2x = 12 - x? Yes, x = 4

Can 2x + 3 = 12 - x? Yes, x = 3

Can x + 3 = 12 - x? No, this reduces to 2x = 9. We have to have integer x

Can x + 6 = 12 - x? Yes, x = 3

Can x + 8 = 12 - x? Yes, x = 2

Therefore, the answer is C

 

I understand how ncr(n,x) = ncr(n,n-x).

I'm not following what you're testing for each multiple choice answer.

Could you explain? I'd like to learn from this.

[nanin]

1. (d)
   
2. (b)
   
3. (b)
   
4. x! / (x-n)!n! and (x+1)! / (n+1)!(x-n)!
   
5. (a)
   

[nanin]

1. (d)
   
2. (b)
   
3. (b)
   
4. x! / (x-n)!n! and (x+1)! / (n+1)! (x-n)!
   
5. (a)
   

[TexasGuy]

1. 5C3+5C3/10C3

2. 20/12P3 [A-J=10 possibility and reversely another 10 possibilities]

3. 10P2*8P2*6P2* 4P2 * 2P2= 10!

4. XCn X+1Cn+1

n+1 X+1

obviously X+1>n+1, So, Ans is B

5.12Cx+p=12Cx

or, 12=x+p+x=2x+p

putting the value of P in the 1st relation, All the value satisfy the relation except, © P=3

 

 

I dont understand number 5

[TexasGuy]

nevermind...i get it now

[mustakim]

For the question number 2, I have some confusion. From my point of view, the answer should be 1/22. Here is the reason why:

A - L have 12 letters. From here, 10 possible case to choose 3 consecutive letters.

12 letters could be arranged in 12*11*10 ways. When each 3 consecutive letters could be arranged 6, which in total 6*10.

So, the answer will be (6*10) / (12*11*10) = 1/22.

I am not so much confident about my answer, but I think this should be correct.