Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 22 Oct 2016, 00:16

### 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

# If there are ten positive real numbers n1 < n2 < n3 … < n10, how many

Author Message
TAGS:

### Hide Tags

Manager
Joined: 08 Oct 2010
Posts: 213
Location: Uzbekistan
Schools: Johnson, Fuqua, Simon, Mendoza
WE 3: 10
Followers: 9

Kudos [?]: 607 [0], given: 974

If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

### Show Tags

11 Nov 2010, 04:02
2
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

56% (02:26) correct 44% (01:23) wrong based on 25 sessions

### HideShow timer Statistics

If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a) 45
b) 90
c) 120
d) 180
e) 150

One solving method is the following:
Three numbers can be selected and arranged out of ten numbers in 10P3 ways=10!/7!=10*9*8. Now this arrangement is restricted to a given condition that first number is always less than the second number, and the second number is always less than the third number. Hence three numbers can be arranged among themselves in 3! ways.

Required number of arrangements=(10*9*8)/(3*2)=120

(the source: Winners’ Guide to GMAT Math – Part II)

Pls, can someone explain me, why 10P3 is divided by 3! ? Inasmuch as I understand the denominator denotes that arragements cases of repeated numbers are to be excluded. Take we this into account then 3! in the denominator has to do with cases such as (n1, n1, n1) or (n2, n2, n2) … But, the question asked in this problem is of a different kind. Then what does 3! mean or has 3! actully to do with triplets with repeated numbers which I could not comprehend?

Many thanks beforhand for detailed explanations !!!
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 35242
Followers: 6619

Kudos [?]: 85350 [3] , given: 10236

Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

### Show Tags

11 Nov 2010, 04:15
3
KUDOS
Expert's post
feruz77 wrote:
If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a) 45
b) 90
c) 120
d) 180
e) 150

One solving method is the following:
Three numbers can be selected and arranged out of ten numbers in 10P3 ways=10!/7!=10*9*8. Now this arrangement is restricted to a given condition that first number is always less than the second number, and the second number is always less than the third number. Hence three numbers can be arranged among themselves in 3! ways.

Required number of arrangements=(10*9*8)/(3*2)=120

(the source: Winners’ Guide to GMAT Math – Part II)

Pls, can someone explain me, why 10P3 is divided by 3! ? Inasmuch as I understand the denominator denotes that arragements cases of repeated numbers are to be excluded. Take we this into account then 3! in the denominator has to do with cases such as (n1, n1, n1) or (n2, n2, n2) … But, the question asked in this problem is of a different kind. Then what does 3! mean or has 3! actully to do with triplets with repeated numbers which I could not comprehend?

Many thanks beforhand for detailed explanations !!!

Consider the following approach: we can choose $$C^3_{10}=120$$ different triplets out of 10 distinct numbers. Each triplet (for example: {a,,b,c}) can be arranged in 3! ways ({a,b,c}; {a,c,b}, {b,c,a}, ...), but only one arrangement (namely {a,b,c}) will be in ascending order, so basically $$C^3_{10}=120$$ directly gives the desired # of such triplets.

If you look at the approach you posted: $$\frac{P^3_{10}}{3!}$$ then you can notice that it's basically the same because: $$\frac{P^3_{10}}{3!}=C^3_{10}=120$$

Similar problems:
probability-of-picking-numbers-in-ascending-order-89035.html?hilit=ascending%20probability
probability-for-consecutive-numbers-102191.html?hilit=ascending%20probability#p793614

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12170
Followers: 538

Kudos [?]: 151 [0], given: 0

Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

### Show Tags

29 Aug 2016, 09:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many   [#permalink] 29 Aug 2016, 09:29
Similar topics Replies Last post
Similar
Topics:
1 If a and b are both odd prime numbers and a < b, then how many differe 6 10 Jun 2016, 05:09
22 How many integers N are prime numbers in the range 200 < N < 220? 18 21 Apr 2016, 13:45
22 How many prime numbers n exist such that 90 < n < 106 and 3 11 Jun 2013, 03:59
11 How many prime numbers n exist such that 90 < n < 106 and n 5 01 May 2013, 22:37
17 How many integers n are there such that 1< 5n +5 < 25? 16 24 Sep 2012, 05:04
Display posts from previous: Sort by