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permutation ques

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Manager
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permutation ques [#permalink] New post 27 Oct 2007, 05:24
Hi Guys,

Can anyone help me in the following ques ?

Q- What is the SUM of all 3 digit numbers that can be formed using the digits 1,2 & 3 ? (repetition of digits not allowed)

(how can one calc sum)

Thanks,
JG
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Re: permutation ques [#permalink] New post 27 Oct 2007, 06:37
greatchap wrote:
Hi Guys,

Can anyone help me in the following ques ?

Q- What is the sum of all 3 digit numbers that can be formed using the digits 1,2 & 3 ? (repetition of digits not allowed)

(how can one calc sum)

Thanks,
JG


All possible numbers out of 1,2,3 = 3! = 6 numbers
3 will appear TWICE (at unit, tenth & hundreth place) same for 2 & 1
Summation of Units place= 2X3 + 2X2 + 2X1 = 12
Summation of Tens place=1+12=13
Summation of 100th place=1+12=13

Summation = 1332
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 [#permalink] New post 27 Oct 2007, 10:59
123
132
213
231
312
321
-----
1332



just write them out and add... much faster (20 seconds) vs amit's way of 100 seconds
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 [#permalink] New post 27 Oct 2007, 11:19
StartupAddict wrote:
123
132
213
231
312
321
-----
1332



just write them out and add... much faster (20 seconds) vs amit's way of 100 seconds


Hahaha.....good one ;-)

Dude now let's extend this challenge...........instead of 1,2,3 do it for 1,2,3,4,5..........i suppose on GMAT we can expect a little trickier than 1,2,3......... ;-).........now what????? Tell me???? i'm rite here ;-)
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 [#permalink] New post 27 Oct 2007, 11:32
singh_amit19 wrote:
StartupAddict wrote:
123
132
213
231
312
321
-----
1332



just write them out and add... much faster (20 seconds) vs amit's way of 100 seconds


Hahaha.....good one ;-)

Dude now let's extend this challenge...........instead of 1,2,3 do it for 1,2,3,4,5..........i suppose on GMAT we can expect a little trickier than 1,2,3......... ;-).........now what????? Tell me???? i'm rite here ;-)


singh_amit19

Your method is valid

1,2,3,4,5

n = 5! = 120

120/5 = 24

24*1 = 24
24*2 = 48
24*3 = 72
24*4 = 96
24*5 = 120

total = 360

360000+36000+3600+360+36 = 399,996

:)
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kool [#permalink] New post 27 Oct 2007, 20:44
Thanks a lot guys for the solution.

I came across a similar ques. can anyone explain this one.

Q- How many number of times will the digit 3 be written when listing integers from 1 to 1000?
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 [#permalink] New post 27 Oct 2007, 21:28
I learned, and had to relearn how to do your final question.

how many times does 3 appear in 1-1000?
number of times 3 appears all 3 times: 333 = 3
number of times 3 appears as 2 digits:
33x, 3x3, xx3
Probablities of each digit space for 33x are:
(1)(1)(9)=9, multiply by 3 to get the three different digit sets = 27 numbers that have three, but three appears twice each set, so 27*2=54 3's.
number of times 3 appears as 1 digit:
3xx, x3x, xx3
Probabilities:(1)(9)(9)=81, times three for the three sets = 243 (don't multiply this again because 3 appeared on once each set)
so, 243+3+54=300

Hope that helps.
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kool...thx [#permalink] New post 29 Oct 2007, 03:52
[quote="ben928"]I learned, and had to relearn how to do your final question.

how many times does 3 appear in 1-1000?
number of times 3 appears all 3 times: 333 = 3
number of times 3 appears as 2 digits:
33x, 3x3, xx3
Probablities of each digit space for 33x are:
(1)(1)(9)=9, multiply by 3 to get the three different digit sets = 27 numbers that have three, but three appears twice each set, so 27*2=54 3's.
number of times 3 appears as 1 digit:
3xx, x3x, xx3
Probabilities:(1)(9)(9)=81, times three for the three sets = 243 (don't multiply this again because 3 appeared on once each set)
so, 243+3+54=300

Hope that helps.[/quote]

Hey, [b]thanks[/b] for the explaination...I understood the logic. Man does these type of ques really come. :roll:
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new ques [#permalink] New post 29 Oct 2007, 04:03
this new question confused me ...

Q- How many numbers greater than 0 and less than one million can be formed with the digits 0,7,8 ?
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 [#permalink] New post 30 Oct 2007, 05:19
[quote="KillerSquirrel"][quote="singh_amit19"][quote="StartupAddict"]

singh_amit19

Your method is valid

1,2,3,4,5 ~ abcde

n = 5! = 120

120/5 = 24 ~ [color=red]Does it mean for each place (ie: a,b,c,d,e) each digit appear 24 times?[/color]
24*1 = 24
24*2 = 48
24*3 = 72
24*4 = 96
24*5 = 120

total = 360

360000+36000+3600+360+36 = 399,996 ~ [color=red]why was it 36 + 360+3600... Why didnt it start with 360 as you caculated total above?[/color]

:)[/quote]

This is such an interesting way!! I didnt quite understand your caculation. Please kindly help explain :)!
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difference [#permalink] New post 03 Nov 2007, 07:11
Can anyone tell me the difference between the two ques:

Q-1 How many six letter words can be formed by re-arranging the letters of the word 'NATIVE' begining with 'N' and ending with 'E'.

A- here we did 4P4. or 4! as the two words r fixed.
If they were not fixed we would have multiplied by 2!. OK.

Q-2 In how many ways can the letters of the word 'DOUBLE' be re-arranged such that the order in which the vowels appear in the word does not change.

A- Here shouldnt it be _OU_ _ E. so OUE fixed. so why isnt anwer 3!. or if its isnt fixed why isnt answer 3! * 3! .

Thanks,
Please answer my query.
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Re: difference [#permalink] New post 03 Nov 2007, 07:20
greatchap wrote:
Can anyone tell me the difference between the two ques:

Q-1 How many six letter words can be formed by re-arranging the letters of the word 'NATIVE' begining with 'N' and ending with 'E'.

A- here we did 4P4. or 4! as the two words r fixed.
If they were not fixed we would have multiplied by 2!. OK.

Q-2 In how many ways can the letters of the word 'DOUBLE' be re-arranged such that the order in which the vowels appear in the word does not change.

A- Here shouldnt it be _OU_ _ E. so OUE fixed. so why isnt anwer 3!. or if its isnt fixed why isnt answer 3! * 3! .

Thanks,
Please answer my query.


How many six letter words can be formed by re-arranging the letters of the word 'NATIVE' begining with 'N' and ending with 'E'.

Since N and E are constant then you are asking about rearranging four letters ---> 4!

In how many ways can the letters of the word 'DOUBLE' be re-arranged such that the order in which the vowels appear in the word does not change.

Same solution here you need to rearrange three letters ---> 3! ---> but you can also move some vowels around (i.e _OU_ _ E or OU_ _ E_ or OU__ _ E) just as long you keep them in the original order.

:)
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 [#permalink] New post 03 Nov 2007, 07:48
alohagirl wrote:
KillerSquirrel wrote:
singh_amit19 wrote:
StartupAddict wrote:

singh_amit19

Your method is valid

1,2,3,4,5 ~ abcde

n = 5! = 120

120/5 = 24 ~ Does it mean for each place (ie: a,b,c,d,e) each digit appear 24 times?

24*1 = 24
24*2 = 48
24*3 = 72
24*4 = 96
24*5 = 120

total = 360

360000+36000+3600+360+36 = 399,996 ~ why was it 36 + 360+3600... Why didnt it start with 360 as you caculated total above?

:)


This is such an interesting way!! I didnt quite understand your caculation. Please kindly help explain :)!


the logic is simple:

What is the sum of all 3 digit numbers that can be formed using the digits 1,2 & 3 ? (repetition of digits not allowed).

To find all the 3 digit numbers we apply 3! = 6

123
132
213
231
321
312

since we used 3 different digits (i.e. 1,2,3) then we have two numbers that start with 1, two numbers that start with 2, and two numbers that start with 3 ---> 6/3 = 2 --> for your above question the answer is yes !

1*2 = 2
2*2 = 4
3*2 = 6

total = 12

12*100+12*10+12*1 = 1332 ---> I started with 12*100 since we had three digit numbers (i.e. base 100)

you can apply the same logic for five digit as well.

:)
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Re: permutation ques [#permalink] New post 03 Nov 2007, 08:03
greatchap wrote:
Hi Guys,

Can anyone help me in the following ques ?

Q- What is the sum of all 3 digit numbers that can be formed using the digits 1,2 & 3 ? (repetition of digits not allowed)

(how can one calc sum)

Thanks,
JG


3! ways to arrange the digist w/ no repitition.

123
132
231
213
321
312

1332
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Re: [#permalink] New post 07 Mar 2008, 04:13
KillerSquirrel wrote:
alohagirl wrote:
KillerSquirrel wrote:

singh_amit19

Your method is valid

1,2,3,4,5 ~ abcde

n = 5! = 120

120/5 = 24 ~ Does it mean for each place (ie: a,b,c,d,e) each digit appear 24 times?

24*1 = 24
24*2 = 48
24*3 = 72
24*4 = 96
24*5 = 120

total = 360

360000+36000+3600+360+36 = 399,996 ~ why was it 36 + 360+3600... Why didnt it start with 360 as you caculated total above?

:)


hey if we apply the same logic for numbers 1,2,3,4 & 5 as mentioned above then wont the answer be - 3999960

24*1 = 24
24*2 = 48
24*3 = 72
24*4 = 96
24*5 = 120

total = 360

So 360 + 360*10 + 360*100 + 360*1000 + 360* 10000
=> 360 + 3600 + 36000 + 360000 + 3600000
= 3999960

AM I right ?
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Re: [#permalink] New post 07 Mar 2008, 07:31
StartupAddict wrote:
123
132
213
231
312
321
-----
1332



just write them out and add... much faster (20 seconds) vs amit's way of 100 seconds



yea, theres only 3! = 6 sets to write out. would be very quick.
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Re: permutation ques [#permalink] New post 07 Mar 2008, 08:18
Permutation & Combination is a topic which has always bothered me.

Here are 3 ques which I am unable to solve.

Q-1 A committe of 2 hawkers and 3 shopkeepers is to be formed from 7 hawkers and 10 shopkeepers. Find the number of ways in which this can be done if a particular shopkeeper is included and a particular hawker is excluded ?

(a) 120 (b) 240 (c) 540 (d) 720 (e) 2520

Q-2 In how many ways can a committe of 5 persons be formed out of 6 men and 4 women when at least one women has to be necessarily selected ?

(a) 6 (b) 252 (c) 246 (d) 251 (e) None of these

Q-3 During a tournament, each of the ten members of a certain chess club plays every other member exactly three times. How many games occur during the tournament ?

(a) 3^10 (b) 10^3 (c) 1200 (d) 135 (e) 720

[The ques below was solved by me but I have a query]
Q-X Everyone shakes hand with everyone. If there are 66 handshakes whats the total number of people.

(a) 11 (b) 12 (c) 13 (d) 14 (e) 15

Answer is : 12.. I did it. What I did was put values one by one. Like 11C2 or 12C2 and see if its 66 or not.

But what I wanna ask is why we do 12C2. I mean why C2. Whats the importance of 2 here.

Can anyone provide explanations as well. Please. Thanks a lot.
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Re: permutation ques [#permalink] New post 07 Mar 2008, 09:39
A committee of 2 hawkers and 3 shopkeepers is to be formed from 7 hawkers and 10 shopkeepers. Find the number of ways in which this can be done if a particular shopkeeper is included and a particular hawker is excluded?

This problem was a bit messy because you have to be sure of what exacly you are doing. Took me about 3 minutes to clean up my own thoughts.

7 hawkers
Exclude 1 --> 6 hawkers
6C2 = 15

10 shopkeepers
If we include one in the group of 3, we only have to choose 2 to be in the group of 3.
We have to account for the one that we already included. 10 - 1
9C2 = 36

15*36 = 540
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Re: permutation ques [#permalink] New post 07 Mar 2008, 09:47
In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one women has to be necessarily selected ?
At least one = Total - none

Total
10C5 = 252

None
6C5 = 6

252-6 = 246
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Re: permutation ques [#permalink] New post 07 Mar 2008, 09:50
During a tournament, each of the ten members of a certain chess club plays every other member exactly three times. How many games occur during the tournament ?

plays each other = 10C2
plays each other twice = 2* 10C2
plays each other 50 times = 50 *10c2


plays each other 3 times = 3 *10C2 = 135
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Re: permutation ques   [#permalink] 07 Mar 2008, 09:50
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