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Re: permutation ques [#permalink]
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
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
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
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
[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.
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.
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.
Re: permutation ques [#permalink]
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.
Re: permutation ques [#permalink]
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
Re: permutation ques [#permalink]
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
Re: permutation ques [#permalink]
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 _________________
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
gmatclubot
Re: permutation ques
[#permalink]
07 Mar 2008, 09:50