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

 It is currently 11 Dec 2013, 02:11

# Events & Promotions

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

# In how many ways can 3-digit numbers be formed selecting 3 d

Author Message
TAGS:
Manager
Joined: 24 Sep 2008
Posts: 194
Schools: MIT / INSEAD / IIM - ABC
Followers: 1

Kudos [?]: 10 [0], given: 7

In how many ways can 3-digit numbers be formed selecting 3 d [#permalink]  16 Aug 2009, 03:13
00:00

Difficulty:

5% (low)

Question Stats:

100% (02:27) correct 0% (00:00) wrong based on 2 sessions
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!
Intern
Joined: 10 Jul 2009
Posts: 48
Followers: 1

Kudos [?]: 5 [0], given: 4

Re: P&C - 3-digit numbers..... [#permalink]  19 Aug 2009, 20:39
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!

My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.
Manager
Joined: 14 Aug 2009
Posts: 124
Followers: 2

Kudos [?]: 86 [1] , given: 13

Re: P&C - 3-digit numbers..... [#permalink]  19 Aug 2009, 22:45
1
KUDOS
tomirisk wrote:
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!

My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.

can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

_________________

Kudos me if my reply helps!

Intern
Joined: 10 Jul 2009
Posts: 48
Followers: 1

Kudos [?]: 5 [0], given: 4

Re: P&C - 3-digit numbers..... [#permalink]  20 Aug 2009, 03:40
flyingbunny wrote:
tomirisk wrote:
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!

My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.

can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

you are right flyingbunny that we can't discard a double digit. But answer C=4^3=64
and your solution 5P3-3(3P2+3P1)= 34. how can it be C?
Manager
Joined: 14 Aug 2009
Posts: 124
Followers: 2

Kudos [?]: 86 [0], given: 13

Re: P&C - 3-digit numbers..... [#permalink]  20 Aug 2009, 04:16
ha, 5P3-3(3P2+3P1)=60-3*9=33
_________________

Kudos me if my reply helps!

Senior Manager
Joined: 10 Dec 2008
Posts: 484
Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
Followers: 21

Kudos [?]: 86 [0], given: 12

Re: P&C - 3-digit numbers..... [#permalink]  20 Aug 2009, 12:25
Umm... you want to explain your solution for laypeople?
Intern
Joined: 14 Aug 2009
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 18

Re: P&C - 3-digit numbers..... [#permalink]  21 Aug 2009, 10:20
flyingbunny wrote:
tomirisk wrote:
GODSPEED wrote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!

My answer is 4P3. We should discard a double digit. We only have 4 distinctive numbers to choose from.

can't discard duplicated digits, for instance, the number can be 121.

the right one should be 5P3-3(3P2+3P1)=33

the SOL looks right.
Manager
Joined: 12 Aug 2009
Posts: 107
Followers: 3

Kudos [?]: 12 [0], given: 2

Re: P&C - 3-digit numbers..... [#permalink]  22 Aug 2009, 06:10
Senior Manager
Joined: 22 Dec 2009
Posts: 366
Followers: 9

Kudos [?]: 167 [0], given: 47

Re: P&C - 3-digit numbers..... [#permalink]  05 Jan 2010, 13:12
flyingbunny wrote:
ha, 5P3-3(3P2+3P1)=60-3*9=33

Can you please explain the solution...

Thanks,
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Manager
Joined: 25 Aug 2009
Posts: 178
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 6

Kudos [?]: 85 [1] , given: 3

Re: P&C - 3-digit numbers..... [#permalink]  05 Jan 2010, 18:42
1
KUDOS
For those of you who need a detailed explanation
Lets find the number of 3 digit numbers with only single 1.
i.e. 3 digit numbers from 1,2,3,4
i.e. 4P3 (since order is important)

Next we find numbers with two 1s in it.
the third digit can be chosen in 3C1 ways and the numbers can be arranged in !3 ways, but 2 numbers are the same hence we divide by !2 (Formula used - number of ways of arranging p things which have q things of one kind, r things of another kind and so on is !p/(!q+!r+...))
Finally we have 3C1*(!3/!2))

Hope this helps
_________________

Rock On

Senior Manager
Joined: 22 Dec 2009
Posts: 366
Followers: 9

Kudos [?]: 167 [0], given: 47

Re: P&C - 3-digit numbers..... [#permalink]  06 Jan 2010, 11:18
Thanks Atish!
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Manager
Joined: 27 Aug 2009
Posts: 152
Followers: 1

Kudos [?]: 8 [0], given: 1

Re: P&C - 3-digit numbers..... [#permalink]  06 Jan 2010, 11:28
Thanks for the explanation
Intern
Joined: 20 Dec 2009
Posts: 14
Followers: 1

Kudos [?]: 10 [0], given: 5

Re: P&C - 3-digit numbers..... [#permalink]  08 Jan 2010, 04:23
I've a doubt regarding the answer and explanation given.
The question does not mention any constrain on the repetition of digits, so we can have 4 distinct choices for each position of the 3-digited number and hence we have
4^3 = 64 such numbers.
Can any one explain why this can't be the answer?

TIA.
Manager
Joined: 05 Jul 2008
Posts: 139
GMAT 1: Q V
GMAT 2: 740 Q51 V38
Followers: 2

Kudos [?]: 48 [0], given: 40

Re: P&C - 3-digit numbers..... [#permalink]  09 Jan 2010, 06:13
ashueureka wrote:
I've a doubt regarding the answer and explanation given.
The question does not mention any constrain on the repetition of digits, so we can have 4 distinct choices for each position of the 3-digited number and hence we have
4^3 = 64 such numbers.
Can any one explain why this can't be the answer?

TIA.

Quote:
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

It means you just have 1 digit 2, 1 digit 3, 1 digit 4 and 2 digits 1 to form a number.
Senior Manager
Joined: 02 Aug 2009
Posts: 269
Followers: 3

Kudos [?]: 73 [0], given: 1

Re: P&C - 3-digit numbers..... [#permalink]  09 Jan 2010, 08:46
hi expl of correct ans..
there are four different digits so no of ways 3 digits can be chosen... 4P3...
rest 3 digit nos will have two 1's and a digit from rest 3.... 3!/2!(noof ways when two digits are same)*3C1(3 different digits)... so ans is 4P3+3!/2!*3C1...D
Re: P&C - 3-digit numbers.....   [#permalink] 09 Jan 2010, 08:46
Similar topics Replies Last post
Similar
Topics:
How many even numbers of 3 digits can be formed with the 10 04 Jul 2004, 21:17
How many different 3 digit possible integers can be formed 4 22 Aug 2004, 19:08
How many 3 digit odd numbers can be formed from the set of 7 13 Jan 2005, 05:22
What is the sum of the 3 digit numbers that can be formed 5 02 Jun 2005, 02:52
2 In how many ways can 3-digit numbers be formed selecting 3 d 14 16 Aug 2009, 03:13
Display posts from previous: Sort by