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
(C) 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
(C) 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
(C) 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
(C) 3
(D) 6
(E) 8

[kamakshi]

1 [D]
2 [B]
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 [B] 4 [B] 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, (C) 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 [B] would be the answer.

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