Hi mandar,the answer to the second question:nC2 = 45 => n!/(2!)(n-2)! = n(n-1) = 90 => n(n-1) = 10X 9so, n=no. of teams = 10.bye;wrote:*mandar keny <mandar_gec@...>*`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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- 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 competetionname the team as x1 , x2 , x3 ...........xnx1 will play n-1 gamesx2 will also play n-1 games but we have already included match between x1 and x2 above.so it n-2x3 ....... n-3x4 ......... n-4......xn .......n-nnow add them(n-1) + (n-2) + (n-3)........(n-n) = 45 (given 45)(n+n+n......n times) - (1 + 2 + 3 + 4.......n) =45n^2 - (n(n+1)/2) = 452*n^2 - n^2 - n = 90n^2 - n -90 = 0(n - 10)*(n + 9) = 0therefore n = 10 (because n cannot be negative)is it k now???
wrote:*suvarchalarani <raninistala@...>***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 Mandar

If u got the explanation to this as to how you arrive at nC2 then, fine.

Otherwise, think on these lines

A game could be played between any 2 teams, right?

when there are n teams , everytime 2 teams are selected in nC2 ways

given that 45 games are played

so, nC2=45

If 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=10

Or by lenghty method, solve the quadratic equation that results from

the above as:

n^2-n-90=0

Discard the negative value for n and consider the positive value only

Hope this explanation helped you

Good luck

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