official quant thread for cat08

[srikar2097]

Originally Posted by asghan

Yup thats the answer given, but can't we say that 8C4(min batsman)*6C4(min bowler)*2C1(Wicketkeepr)*6C3(6 players will be left from batsmen and bowler after selecting 4 each, and we need 3 more to complete the team) ???

But by this method, how would you say exactly how many batsmen are there and how many bowlers are there? I guess that's the mistake.

[the_hate]

Quote:
Originally Posted by srikar2097
Are you sure answer is 1652?

Here's my approach:
Wicketkeeper = 2C1 = 2 ways
Batsmen = 8C4 = 70 ways
So 5 players are fine. Remaining 6 players are all bowlers. i.e. 1 way

So total=2*70=140 ways

aren't u missin out some cases...thr can be a case with 1 WC,5batsman and 5 bowlers or 1 WC, 6Batsman and 4 bowlers....

Originally Posted by asghan

Yup thats the answer given, but can't we say
that 8C4(min batsman)*6C4(min bowler)*2C1(Wicketkeepr)*6C3(6 players will be left from batsmen and bowler after selecting 4 each, and we need 3 more to complete the team) ???

this seems to be wrong method...
in this case there will be repetition of cases...
say u select sachin as batsman in first 4(min batsman) and select sehwag in 6 players left...
one more case wud be when u select sehwag in first min 4 batsman...and sachin in 6 remainin players...
all other players remainin the same...
but the team formed from these two ways wud essentially be the same..
hence thr will be sm xtra cases in this method...

cheers

[asghan]

Originally Posted by srikar2097

But by this method, how would you say exactly how many batsmen are there and how many bowlers are there? I guess that's the mistake.

The constraint was atleast 4 batsman atleast 4 bowlers and 1 wicketkeeper so after selecting atleast 4 bowlers in 6C4 ways and 4 batsman in 8C4 ways we can coose the rest 3 players out of the 6 remaining players in any which way dosent matter whether they are bowler or batsmen.

I had this thing in mind while solving it.

please shed more light??

[romy4allcse]

Originally Posted by srikar2097

Are you sure answer is 1652?

Here's my approach:
Wicketkeeper = 2C1 = 2 ways
Batsmen = 8C4 = 70 ways
So 5 players are fine. Remaining 6 players are all bowlers. i.e. 1 way

So total=2*70=140 ways

A cricket team of 11 members is to be selected from 8 batsman 6 bowlers and 2 wicketkeeper, in how many ways 11 members can be selected such that there are atleast 4 batsman, 4 bowlers and only one wicketkeepr ??

given we hav to chose atleast 4 batsman and 4 bowlers out of 8 and 6 respectively.
so no of ways chosing 11 can be done in these ways
4(batsman) + 6(bowlers) + 1(wck)
8c4*6c6* 2c1 = 140
5(batsman) + 5(bowlers) + 1(wck)
8c5*6c5*2c1 = 672
6(batsman) + 4(bowlers) + 1(wck)
8c6*6c4*2c1 = 840
hence total no of ways of selecting 11 players = 140 + 672 + 840
= 1652

[srikar2097]

Originally Posted by The Last Romeo

i guess u r right, the method suggested by u was short and sweet. But just out of curiosity i wanted to know on how do we perform sythentic divison, if you can throw some light on it thaen it would be great.
Also were u able to attmept my last question??

Reg Q3. For a bi-quad equation.
Sum of roots = -b/a
product of roots = e/a
sum of product of roots taken 3 at a time = d/a
sum of product of roots taken 2 at a time = c/a

Expand the eq. Apply this and you get choice (b)

Reg. Synthetic division - follow this link

[raghav507]

Originally Posted by The Last Romeo

Hi,
If not more trouble., could please elaborate on the approach followed.

sure buddy,
thoda sa typing main mushkil hai but still yeh lo...

let the roots are p, q, r, s (only for representational purpose)
from the given eqn we get...
pqrs = e/a
pqr+qrs+rsp+spq = -d/a
pq+qr+rs+sp+rp+qs = c/a
p+q+r+s = -b/a

the given expression is
(1-s1)(1-s2)(1-s3)(1-s4)
= 1 - (s1 + s2 + s3 + s4)
+ (s1s2 + s2s3 + s3s4 + s4s1 + s2s4 + s1s3)
- (s1s2s3 s2s3s4 s3s4s1 s1s2s4)
+ (s1s2s3s4)
= 1 - (-b/a) + (c/a) - (-d/a) + (e/a)
= (a+b+c+d+e)/a

hence the answer...
hope it helps...

[raghav507]

Originally Posted by srikar2097

Reg Q3. For a bi-quad equation.
Sum of roots = -b
product of roots = e
sum of product of roots taken 3 at a time = d
sum of product of roots taken 2 at a time = c

Expand the eq. Apply this and you get choice (b)

Reg. Synthetic division - follow this link

hey buddy,
you forgot to use the coefficient 'a' in your post...
it will be -b/a, e/a, -d/a, c/a respectively...

[prakharc]

There are 3 possible cases

Case 1:- 1 W, 6 Batsman,4 Bowlers
No. of ways= 2*8C6*6C4=840

Case 2:- 1W, 5 Batsman, 5 Bowlers
No. of ways= 2* 8C5 * 6C5 = 672

Case 3:- 1W, 4Batsman, 6 Bowlers
No. Of ways= 2* 8C4* 6C6 = 140

Total number of ways = 840+672+140=1652 ways Ans

[srikar2097]

Originally Posted by the_hate

[/i]

aren't u missin out some cases...thr can be a case with 1 WC,5batsman and 5 bowlers or 1 WC, 6Batsman and 4 bowlers....


this seems to be wrong method...
in this case there will be repetition of cases...
say u select sachin as batsman in first 4(min batsman) and select sehwag in 6 players left...
one more case wud be when u select sehwag in first min 4 batsman...and sachin in 6 remainin players...
all other players remainin the same...
but the team formed from these two ways wud essentially be the same..
hence thr will be sm xtra cases in this method...

cheers

Yup, I jumped the gun here! I guess was in a hurry to have lunch
Thanks...

[srikar2097]

Originally Posted by raghav507

hey buddy,
you forgot to use the coefficient 'a' in your post...
it will be -b/a, e/a, -d/a, c/a respectively...

Yup! Thanks. Have edited my original post too..