1) if a monkey is given 11 letters that can form the word : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 14:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# 1) if a monkey is given 11 letters that can form the word

Author Message
Intern
Joined: 04 Feb 2004
Posts: 28
Location: USA
Followers: 0

Kudos [?]: 5 [0], given: 0

1) if a monkey is given 11 letters that can form the word [#permalink]

### Show Tags

25 Apr 2005, 11:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1) if a monkey is given 11 letters that can form the word "probability", what is the probability that the monkey will form the word "probability"?

2)A player selects six numbers from 1 to 49. What is the probability of picking the six winning numbers?

3) A typewriter has 48 keys (including space, numbers, etc). What is the probability that a baby, who strikes keys randomly, will correctly type the word "hamlet"?
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [0], given: 0

### Show Tags

25 Apr 2005, 11:41
1) 1/11!
2) 6!/(49*48*47*46*45*44)
3) 1/48^6

will explain if correct...
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [0], given: 0

### Show Tags

25 Apr 2005, 12:20
1) 1/(11!/2!*2!)

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Manager
Joined: 05 Feb 2005
Posts: 116
Location: San Jose
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

25 Apr 2005, 13:02
1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

_________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.

VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [0], given: 0

### Show Tags

25 Apr 2005, 13:08
mckenna wrote:
1) 1/(11!/2!*2!) = 4/11!
Correct.

2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

1/49C6 = 6! x 43! / 49! = 6! / (49x48x47x46x45x44)

oh right ! too late here in germany you can approach this problem with combination- as well as with permutation-reasoning...
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Intern
Joined: 04 Feb 2004
Posts: 28
Location: USA
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

25 Apr 2005, 13:53
Can you guys explain 1?
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 327 [0], given: 0

### Show Tags

25 Apr 2005, 14:19
rc1979 wrote:
Can you guys explain 1?

divide by 2! and 2! because there are 2 bs and 2 s. for example probabilty with the first b in 4th position and the second b in 6th position is 1 way and when you change the b`s you get a second way, but it is the same word. so you divide 2.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 45 [0], given: 0

### Show Tags

25 Apr 2005, 18:31
the number 2) was not so easy, but I got 4!/11! and 1/48^6

rc1979, christoph explained it well, there are 2 letters B (le'ts call it B1 and B2 for example)
Usually you have to work with permutation concerning words problems because order normally matters, in that case there will be no difference if you choose B1 or B2, it will be the same word whereas each B is considered as a specific letter. So you'll have twice more chances with B and it's the same idea for I.

finally, normally for a word with 11 different litters the prob will be 1/11! but since you have 2 repetitions of 2 letters, it will be 1*2*2/11!

VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7

Kudos [?]: 98 [0], given: 0

### Show Tags

25 Apr 2005, 19:04
mckenna wrote:
2) 1/(49c6) i assume that order does not matter. 123456=654321. but i think the question is not properly phrased.
Correct.

it is a permutation not a combiation problem. so it is
=1/(49P6)
25 Apr 2005, 19:04
Display posts from previous: Sort by