Official Quant thread for CAT 2013

MBA CAT 2024
21366
@jain4444 said:
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
@scrabbler said:
Options? Kuch to bada sa aata hai, can't calculate value! Something like 24! - 21!*(4! + 20*3!)regardsscrabbler

don't know the correct answer but i got

_ b1 _ b2 _ b3 _ b4 _

when no 2 boys stand together
g1 + g2 + g3 + g4 + g5 = 17
=> 21C4

when 1 pair of boys stand together
g1 + g2 + g3 + g4 = 18
=> 21C3

when 2 pair of boys stand together
g1 + g2 + g3 = 19
=> 21C2

total cases = (21C4 + 21C3 + 21C2)*4!*20!
21367
@amresh_maverick
i did it like this... first select 1 student from each state to satisfy at least criteria. this can be done in 3c1*3c1*3c1=27 ways. other two students can be chosen randomly from remaining lot of 9-3=6 students. and this can be done in 6c2= 15 ways. therefore the answer should be 27*15= 405..
can u help me to find the flaw in this approach?
21368
@catahead said:
25+15+1 = 41
Could u pls explain the method ?
21369
@catahead said:
One from each in 3.3.3 waysThe remaining in 6C2So 27.6C2 = 405Whats the OA
Ismein double counting hai. Should be 108.

regards
scrabbler
21370
@jain4444 said:
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
24! - 4c3*(22!*3!)


21371
@jain4444 said:
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
24! - 4c3*(22!*3!)


21372
@Calvin4ever said:
In how many ways can 2310 be expressed as a product of 3 factors ?? with soln. pls
a * b * c = 2310

2310 = 2 * 3 * 5 * 7 * 11

a = 2^a1 * 3^a2 * 5^a3 * 7^a4 * 11^a5
b = 2^b1 * 3^b2 .......................* 11^b5
c = 2^c1 * ............................... * 11^c5

a1 + b1 + c1 = 1
=> number of soln. = 3

same for others

total number of soln. = 3^5 (ordered pairs)

but we need to find unordered pairs

we have only one case in which any 2 would be same among a , b , c i.e. 1*1*2310 which we have counted as 3!/2 = 3 times

(243 - 3)/3! = 40 + 1 = 41
21373
@jain4444 said:
In how many ways can 4 boys and 20 girls stand in a line so that no 3 of the boys are next to each other?
24! - 4c3*(22!*3!)
21374
@ani6 said:
calculate remainder when 7219485697/72?
49?

72 = 8*9 and the given number is 8k+1 and 9m+4.

regards
scrabbler
21375
@Joey_Sharma said:
@amresh_maverick i did it like this... first select 1 student from each state to satisfy at least criteria. this can be done in 3c1*3c1*3c1=27 ways. other two students can be chosen randomly from remaining lot of 9-3=6 students. and this can be done in 6c2= 15 ways. therefore the answer should be 27*15= 405.. can u help me to find the flaw in this approach?
Suppose the kids are ABC from state 1, DEF from 2 and GHI from 3.

Consider these cases:
1st selection mein you did ADG, second mein you added B and E
1st selection mein you did BDG, second mein you added A and E
1st selection mein you did AEG, second mein you added B and D
1st selection mein you did BEG, second mein you added A and D
You have taken all of these as different cases. However, they are all the same ABDEG case.
Hence double counting...hence flawed.

regards
scrabbler
21376
@jain4444 said:
don't know the correct answer but i got _ b1 _ b2 _ b3 _ b4 _when no 2 boys stand together g1 + g2 + g3 + g4 + g5 = 17 => 21C4when 1 pair of boys stand together g1 + g2 + g3 + g4 = 18 => 21C3

Could also be g1 + g2 + g3 + g5 etc....have not specified which pair are together? Similarly for later case as well...am not convinced :(
@jain4444 said:
when 2 pair of boys stand together g1 + g2 + g3 = 19 => 21C2total cases = (21C4 + 21C3 + 21C2)*4!*20!

regards
scrabbler
21377
@scrabbler said:
Could also be g1 + g2 + g3 + g5 etc....have not specified which pair are together? Similarly for later case as well...am not convinced regardsscrabbler
that's why I multiplied by 4! at the end

if we chose like 4C2*2!*2! = 24 = 4!

so , in both cases we get the same answer
21378
@scrabbler yeah rite....can you provide the elaborate solution...??plzz
21379
@Logrhythm said:
ab ac ad bc bc and cd are 6 diff pairs...a+b = 1a+c = 2b+c = 3this gives a=0,b=1 anc c=2now since a+d=4hence d can only be 4...
how did u conclude that third smallest sum will be b+c only and not a+d ? i didn't get this.
21380
@jain4444 said:
that's why I multiplied by 4! at the end if we chose like 4C2*2!*2! = 24 = 4! so , in both cases we get the same answer
No the 4! only arranges the boys among themselves. Does not decide which of the spaces in between them are empty. In the original; g1, g2, g3, g4, g5, either g2 or g3 or g4 should be 0 and the sum of the rest should be 18 - you have not accounted for that. You have just taken 4 variables and summed them to 18. 3 ways for choosing which one is empty to at the least chahiye.

regards
scrabbler
21381
@ani6 said:
@scrabbler yeah rite....can you provide the elaborate solution...??plzz
That was the elaborate solution....Chinese remainder logic....see the link below for some earlier discussion...

http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-4683551

regards
scrabbler
21382
@Joey_Sharma said:
@amresh_maverick i did it like this... first select 1 student from each state to satisfy at least criteria. this can be done in 3c1*3c1*3c1=27 ways. other two students can be chosen randomly from remaining lot of 9-3=6 students. and this can be done in 6c2= 15 ways. therefore the answer should be 27*15= 405.. can u help me to find the flaw in this approach?
A lot of repetitions

ABC from state X ,MNO from state Y , PQR from state Z

Suppose u selected 1 from each state : A , M and P
and to random, let us say : Q, R

Your group : A,M ,P,Q,R


Now suppose u chose : A, M, R (1 from each state) + P, Q (2 from each state)

Your group : A,M ,P,Q,R ------a repeat

u get the drift ?


21383
@ani6 said:
how many integer values are there for x and y such that 4x+7y=3? where x
y=1 and x=-1 is one of the solution hence generalized solution

x = 7k-1
y = 1-4k

since we want x 7k-1 k
also for yk >= -125
combining the above results we get -125

hence there are 197 possible values of k

ATDH.
21384

@amresh_maverick kya haal chal boss ? CAT 13 ?

-----------------------------------------------------------Itz not a spam 😛

Q. Find the number of pairs possible. x^2-y^2 = 98

21385
@scrabbler said:
No the 4! only arranges the boys among themselves. Does not decide which of the spaces in between them are empty. In the original; g1, g2, g3, g4, g5, either g2 or g3 or g4 should be 0 and the sum of the rest should be 18 - you have not accounted for that. You have just taken 4 variables and summed them to 18. 3 ways for choosing which one is empty to at the least chahiye. regardsscrabbler
hmm did mistake

so , answer should be (21C4 + 21C3*3 + 21C2)*4!*20! right ?