## Re: [Ascent CAT] try to solve-permutation and combination-imp

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• Hi mandar, the answer to the second question: nC2 = 45 = n!/(2!)(n-2)! = n(n-1) = 90 = n(n-1) = 10X 9 so, n=no. of teams = 10. bye; mandar keny
Message 1 of 13 , Mar 3, 2006

Hi mandar,
the answer to the second question:
nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
so, n=no. of teams = 10.
bye;

mandar keny <mandar_gec@...> wrote:
1. in a number consisting of 7 digits i.e 4 zeros and
3 threes. we need to form a 7 digit no. how?
pllzz explain if anyone can... ans is 15

2.in a competition a several teams take part. each
team plays against all opp team. total no games played
is 45. how many teams are there?
ans given is 10
bt hw?

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• Hi mandar the answer to the first question: The first place can be occupied by only 3 since it cannot be 0. 6 places are left now.Here you need to do the
Message 2 of 13 , Mar 5, 2006
Hi mandar
the answer to the first question:
The first place can be occupied by only 3 since it cannot be 0.
6 places are left now.Here you need to do the selection.There are 6 places out of which there are 4 zeroes and 2 threes left.You can do it in 6c2 ways=15.
Hope you understood it.Whenever there is a repetition you should do combination since the order is not important and all of them have to be in the number anways.
OR
First place can be filled in one way.
There are 6 places left with 4 0's and 2 3's.
There is a formula for that
n!/(p!.q!.....)
where 'n' are the places .
P and Q are the  repetitions.

Hi mandar,
the answer to the second question:
nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
so, n=no. of teams = 10.
bye;

mandar keny <mandar_gec@...> wrote:
1. in a number consisting of 7 digits i.e 4 zeros and
3 threes. we need to form a 7 digit no. how?
pllzz explain if anyone can... ans is 15

2.in a competition a several teams take part. each
team plays against all opp team. total no games played
is 45. how many teams are there?
ans given is 10
bt hw?

__________________________________________________________

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• Hi mandar, the ans to ur 1st ques is 1st digit of the 7-digit number cannot be 0, therefore it is 3 for the rest of the 6 positions we have 4 0 s and 2
Message 3 of 13 , Mar 6, 2006
Hi mandar,
the ans to ur 1st ques is
1st digit of the 7-digit number cannot be 0, therefore it is 3
for the rest of the 6 positions we have 4   0's and  2   3's.
therefore, number of combinations is 6!/(4!x2!) = 15

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• Q1 first place can be filled in 1 way ( Only 3 can occupy the plaace) balance 6 places can befilled by 6! ways 0 is repeated 4 times and balance 3 s is
Message 4 of 13 , Mar 6, 2006
Q1

first place can be filled in 1 way ( Only 3 can occupy the plaace)

balance 6 places can befilled by 6! ways

0 is repeated 4 times and balance 3's is reapeted 2 times so

6!/(4!*2!) will be the final answer

Regards

Hi mandar,
the answer to the second question:
nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9
so, n=no. of teams = 10.
bye;

mandar keny <mandar_gec@...> wrote:
1. in a number consisting of 7 digits i.e 4 zeros and
3 threes. we need to form a 7 digit no. how?
pllzz explain if anyone can... ans is 15

2.in a competition a several teams take part. each
team plays against all opp team. total no games played
is 45. how many teams are there?
ans given is 10
bt hw?

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• Hi Mandar, For QUESTION 2 : The formula comes upto : (n*2 - n) / 2 = no of games played. = (n*2 - n) / 2 = 45 = (n - 10) (n + 9) = 45 = n = 10 Regards,
Message 5 of 13 , Mar 9, 2006
Hi Mandar,

For QUESTION 2 :

The formula comes upto  : (n*2  - n) / 2 = no of games played.

=> (n*2  - n) / 2 = 45
=> (n - 10) (n + 9) = 45
=> n = 10

Regards,
Rimpu Varshney

mandar keny <mandar_gec@...> wrote:
1. in a number consisting of 7 digits i.e 4 zeros and
3 threes. we need to form a 7 digit no. how?
pllzz explain if anyone can... ans is 15

2.in a competition a several teams take part. each
team plays against all opp team. total no games played
is 45. how many teams are there?
ans given is 10
bt hw?

__________________________________________________________

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• Hi mandar here is the answer for your question: to form a seven digit number first place should be filled by 3 this can be done in only one way as there are
Message 6 of 13 , Mar 11, 2006
Hi mandar

to form a seven digit number first place should be filled by 3 this can be done in only one way as there are only threes and zeros and the ramaining six places can be filled in 6!/(2!*4!) ways that is the answer is 1*[6!/(2!*4!)] ie 15 ways we have divided 6! by (2!*4!) because as there are 2 threes and 4 zeros left and selecting any three among the two threes or selecting any zero among four zeros is immaterial
rimpu varshney <rimpuv_2000@...> wrote:
Hi Mandar,

For QUESTION 2 :

The formula comes upto  : (n*2  - n) / 2 = no of games played.

=> (n*2  - n) / 2 = 45
=> (n - 10) (n + 9) = 45
=> n = 10

Regards,
Rimpu Varshney

mandar keny <mandar_gec@...> wrote:
1. in a number consisting of 7 digits i.e 4 zeros and
3 threes. we need to form a 7 digit no. how?
pllzz explain if anyone can... ans is 15

2.in a competition a several teams take part. each
team plays against all opp team. total no games played
is 45. how many teams are there?
ans given is 10
bt hw?

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• Hi, Answer for the first question is: since we have 4 zero s and 3 three s we have two distinct numbers(0 and 3) to make a 7 digit number from them we will
Message 7 of 13 , Mar 13, 2006
Hi,
Answer for the first question is:
since we have 4 zero's and 3 three's
we have two distinct numbers(0 and 3)
to make a 7 digit number from them we will require to have
3 to take the first place which can be done in 1way and the
remaining paces will taken by the remaining numbers in 6! ways.
but we have 4 0's and 2 three's to occupy 6 places.hence the total
number of ways are (1*6!)/(2!*4!)=15

--- In ascent4cat@yahoogroups.com, rimpu varshney <rimpuv_2000@...>
wrote:
>
> Hi Mandar,
>
> For QUESTION 2 :
>
> The formula comes upto : (n*2 - n) / 2 = no of games played.
>
> => (n*2 - n) / 2 = 45
> => (n - 10) (n + 9) = 45
> => n = 10
>
> Regards,
> Rimpu Varshney
>
> mandar keny <mandar_gec@...> wrote:
> 1. in a number consisting of 7 digits i.e 4 zeros and
> 3 threes. we need to form a 7 digit no. how?
> pllzz explain if anyone can... ans is 15
>
> 2.in a competition a several teams take part. each
> team plays against all opp team. total no games played
> is 45. how many teams are there?
> ans given is 10
> bt hw?
>
>
>
>
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• mandar has given correct solution and the second question solution is as foolows n combination 2 = 45 ie n!/(n-2)!2!=45 n(n-1)=90 this gives n = 10 ... place
Message 8 of 13 , Mar 18, 2006
mandar has given correct solution and the second question solution
is as foolows
n combination 2 = 45
ie n!/(n-2)!2!=45
n(n-1)=90
this gives n = 10

--- In ascent4cat@yahoogroups.com, phanindra H <phani_acumen@...>
wrote:
>
> Hi mandar
>
> to form a seven digit number first
place should be filled by 3 this can be done in only one way as
there are only threes and zeros and the ramaining six places can be
filled in 6!/(2!*4!) ways that is the answer is 1*[6!/(2!*4!)] ie 15
ways we have divided 6! by (2!*4!) because as there are 2 threes and
4 zeros left and selecting any three among the two threes or
selecting any zero among four zeros is immaterial
> rimpu varshney <rimpuv_2000@...> wrote:
> Hi Mandar,
>
> For QUESTION 2 :
>
> The formula comes upto : (n*2 - n) / 2 = no of games played.
>
> => (n*2 - n) / 2 = 45
> => (n - 10) (n + 9) = 45
> => n = 10
>
> Regards,
> Rimpu Varshney
>
> mandar keny <mandar_gec@...> wrote:
> 1. in a number consisting of 7 digits i.e 4 zeros and
> 3 threes. we need to form a 7 digit no. how?
> pllzz explain if anyone can... ans is 15
>
> 2.in a competition a several teams take part. each
> team plays against all opp team. total no games played
> is 45. how many teams are there?
> ans given is 10
> bt hw?
>
>
>
>
> __________________________________________________________
> Yahoo! India Matrimony: Find your partner now. Go to
>
>
>
>
> ---------------------------------
> Yahoo! Mail
> Use Photomail to share photos without annoying attachments.
>
> Join Ascent's Weekend Classes for CAT 2006 - Starts March 4, 06
>
http://www.ascenteducation.com/india_mba_iim_cat_classes_courses/inte
nsive.shtml
>
>
>
india Business finances B school ranking
>
> ---------------------------------
>
>
> Visit your group "ascent4cat" on the web.
>
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• 2.in a competition a several teams take part. each ... Hi Mandar If u got the explanation to this as to how you arrive at nC2 then, fine. Otherwise, think on
Message 9 of 13 , Mar 22, 2006
`2.in a competition a several teams take part. each > team plays against all opp team. total no games played> is 45. how many teams are there?> ans given is 10> bt hw? Hi MandarIf u got the explanation to this as to how you arrive at nC2 then, fine.Otherwise, think on these linesA game could be played between any 2 teams, right?when there are n teams , everytime 2 teams are selected in nC2 waysgiven that 45 games are playedso, nC2=45If you know by experience, that nC2=45 when n=10, good.Otherwise, nC2=n(n-1)(n-2)!/(n-2)!2!=>n(n-1)=45*2=90=>n(n-1)=10*9=>n=10Or by lenghty method, solve the quadratic equation that results fromthe above as:n^2-n-90=0Discard the negative value for n and consider the positive value onlyHope this explanation helped youGood luck`
• if u don t wanna apply any formula then u can do it by basics......but it takes time.... solution: suppose there are n teams playing in the competetion name
Message 10 of 13 , Mar 30, 2006
if u don't wanna apply any formula then u can do it by basics......but it takes time....
solution:
suppose there are 'n' teams playing in the competetion
name the team as x1 , x2 , x3 ...........xn
x1 will play n-1 games
x2 will also play n-1 games but we have already included match between x1 and x2 above.so it n-2
x3    ....... n-3
x4 ......... n-4
...
...
xn .......n-n
(n-1) + (n-2) + (n-3)........(n-n) = 45   (given 45)
(n+n+n......n times) - (1 + 2 + 3 + 4.......n) =45
n^2 - (n(n+1)/2) = 45
2*n^2 - n^2 - n = 90
n^2 - n -90 = 0
(n - 10)*(n + 9) = 0
therefore n = 10    (because n cannot be negative)

is it k now???

suvarchalarani <raninistala@...> wrote:
```2.in a competition a several teams take part. each> team plays against all opp team. total no games played> is 45. how many teams are there?> ans given is 10> bt hw?Hi MandarIf u got the explanation to this as to how you arrive at nC2 then, fine.Otherwise, think on these linesA game could be played between any 2 teams, right?when there are n teams , everytime 2 teams are selected in nC2 waysgiven that 45 games are playedso, nC2=45If you know by experience, that nC2=45 when n=10, good.Otherwise, nC2=n(n-1)(n-2)!/(n-2)!2!=>n(n-1)=45*2=90=>n(n-1)=10*9=>n=10Or by lenghty method, solve the quadratic equation that results fromthe above
as:n^2-n-90=0Discard the negative value for n and consider the positive value onlyHope this explanation helped youGood luck```

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