1! +2! +3! +4! +…………………..100! = 2k+1
CANT BE WRITTEN AS 2^n * K, n=0
@audiq7
From Statement II:
Either both A and B are negative or both A and B are
positive.
e.g A = 1 and B = 2 or A = –1 and B = –2 etc.
The statement alone is not sufficient to answer
From Statement I:
C –B is greater than A. So nothing can be said on the
basis of this statement alone.
Combining the two statements:
C will be definitely greater than A when A + B both A and B are positive.
But, C may not be greater than A when A + B both A and B are negative.
e.g. A = –1 and B = –2
AB = 2 > 0
A + B = –3.
So, both C = –1.5 and C = 0 satisfies it.
Hence, the question cannot be answered using both
the statements together.
From Statement II:
Either both A and B are negative or both A and B are
positive.
e.g A = 1 and B = 2 or A = –1 and B = –2 etc.
The statement alone is not sufficient to answer
From Statement I:
C –B is greater than A. So nothing can be said on the
basis of this statement alone.
Combining the two statements:
C will be definitely greater than A when A + B both A and B are positive.
But, C may not be greater than A when A + B both A and B are negative.
e.g. A = –1 and B = –2
AB = 2 > 0
A + B = –3.
So, both C = –1.5 and C = 0 satisfies it.
Hence, the question cannot be answered using both
the statements together.
A man is standing at a point P. He can take one step at a time in any of the 4 directions, N-E-W-S. In how many ways can he end up at the same point P after 6 steps
1)256
2)512
3)400
4)200
1)256
2)512
3)400
4)200
@krum said:A man is standing at a point P. He can take one step at a time in any of the 4 directions, N-E-W-S. In how many ways can he end up at the same point P after 6 steps1)2562)5123)4004)200
a,b,c,d are the steps in E-W-N-S dirns..
a+b+c+d = 6
and a=b and c=d ==> a+c=3
0 3; 1 2; 2 1; 3 0
==>2* [6!/(3!^2) + 6!/(2!^2) ] = 2*[20 + 180] = 400....?
and reg nmat, did u also write ur exam on nov9th (11:45am slot) or did u feel the same abt ur slot...?
a+b+c+d = 6
and a=b and c=d ==> a+c=3
0 3; 1 2; 2 1; 3 0
==>2* [6!/(3!^2) + 6!/(2!^2) ] = 2*[20 + 180] = 400....?
and reg nmat, did u also write ur exam on nov9th (11:45am slot) or did u feel the same abt ur slot...?
@krum said:A man is standing at a point P. He can take one step at a time in any of the 4 directions, N-E-W-S. In how many ways can he end up at the same point P after 6 steps1)2562)5123)4004)200
400
H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
@shadowwarrior said:a,b,c,d are the steps in E-W-N-S dirns..a+b+c+d = 6and a=b and c=d ==> a+c=30 3; 1 2; 2 1; 3 0==>2* [6!/(3!^2) + 6!/(2!^2) ] = 2*[20 + 180] = 400....?
yo!, after so long?
Let an unbiased coin is tossed 10 times. Let the probability that head never occur on two consecutive toss is denoted by P.
Here m and n do not have any common factor. Find m Ä‚— n.
OPTIONS
1) 318
2) 720
3) 512
4) 576
@shadowwarrior mine was on 29th, but i do feel the same :mg:
Let an unbiased coin is tossed 10 times. Let the probability that head never occur on two consecutive toss is denoted by P.
Here m and n do not have any common factor. Find m Ä‚— n.
OPTIONS
1) 318
2) 720
3) 512
4) 576
@shadowwarrior mine was on 29th, but i do feel the same :mg:
@astute99 said:H.C.F. of three positive integers €“ A, B and C €“ is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
6a+6b+6c=48
==>a+b+c=8
(5+3-1)c(3-1)
=>7c2=21
==>a+b+c=8
(5+3-1)c(3-1)
=>7c2=21
@krum said:6a+6b+6c=48==>a+b+c=8(5+3-1)c(3-1)=>7c2=21
can you please explain how the equation was turned to (5+3-1)c(3-1)? I'm not aware about this procedure..
@krum said:yo!, after so long?Let an unbiased coin is tossed 10 times. Let the probability that head never occur on two consecutive toss is denoted by P.Here m and n do not have any common factor. Find m — n.OPTIONS1) 318 2) 720 3) 512 4) 576 @shadowwarrior mine was on 29th, but i do feel the same
9*64=576....?
there can be 0 heads, 1 head, 2heads,....5 heads...(in case of 6 or any more heads, its not possible they cant be arranged without 2 of them being separated)
so 6c5, 7c4, 8c3, 9c2, 10c1, 11c0...sum of which is 144...now p - 144/1024 = 9/64 = m/n
after not-so-great CAT & subsequent discussions with people who gave in the same slot as i did, i was prepping for nmat(LR and LS)....now, i thnk i will have to retake the exam coz of LS :|...mugging words from barrons
there can be 0 heads, 1 head, 2heads,....5 heads...(in case of 6 or any more heads, its not possible they cant be arranged without 2 of them being separated)
so 6c5, 7c4, 8c3, 9c2, 10c1, 11c0...sum of which is 144...now p - 144/1024 = 9/64 = m/n
after not-so-great CAT & subsequent discussions with people who gave in the same slot as i did, i was prepping for nmat(LR and LS)....now, i thnk i will have to retake the exam coz of LS :|...mugging words from barrons
@Hypertexter said:can you please explain how the equation was turned to (5+3-1)c(3-1)? I'm not aware about this procedure..
No of integral solution for
x1+x2+x3+...+xr=N is
C(N+r-1,r-1)
x1+x2+x3+...+xr=N is
C(N+r-1,r-1)
@Hypertexter said:can you please explain how the equation was turned to (5+3-1)c(3-1)? I'm not aware about this procedure..
a+b+c=8
number of integral solutions = n+r-1cr-1; think of it as distributing 8 identical objects to 3 persons
here condition is a,b,c>=1
a-1>=0
take a'=a-1
=>a=a'+1
same for b,c
so a+b+c=8
=>a'+1+b'+1+c'+1=8
=>a'+b'+c'=5
now use n+r-1cr-1
@shadowwarrior me too, mugging up from ims wordlist, iift also puts in vocab based ques.
number of integral solutions = n+r-1cr-1; think of it as distributing 8 identical objects to 3 persons
here condition is a,b,c>=1
a-1>=0
take a'=a-1
=>a=a'+1
same for b,c
so a+b+c=8
=>a'+1+b'+1+c'+1=8
=>a'+b'+c'=5
now use n+r-1cr-1
@shadowwarrior me too, mugging up from ims wordlist, iift also puts in vocab based ques.
a and b are natural numbers such that a>b>1.If 8! is divisible by a^2*b^2 ,then how much such sets (a,b) are possible??
a)5
b)6
c)7
d)8
Pls share ur approach,..
@krum said:Let an unbiased coin is tossed 10 times. Let the probability that head never occur on two consecutive toss is denoted by P.Here m and n do not have any common factor. Find m — n.OPTIONS1) 318 2) 720 3) 512 4) 576
ans = (d) 576
Its imp to realize at once that there cannot be more than 5 heads.
so
case 1 : 5 heads = 6c5 =6
case 2 : 4 heads = 7c4 = 35
case 3 : 3 heads = 8c3 = 56
case 4 : 2 heads = 9c2 = 36
case 5 : 1 heads = 10c1 = 10
case 6 : 0 heads = 1
total = 144
possible outcomes = 2^10 (since each of the ten outcomes has 2 possibilities so 2*2*2....)
P = m/n = 144/1024 = 9/64
hence m*n = 9 * 64
ans = (d) 576
NOTE: the probabilities in the 6 cases above are derived and can be understood as filling the alternate gaps wen all the TAILS are first laid out leaving 1 gap in between each of them
| attitude is more important than facts |
@pavimai said:a and b are natural numbers such that a>b>1.If 8! is divisible by a^2*b^2 ,then how much such sets (a,b) are possible??a)5b)6c)7d)8Pls share ur approach,..
8!=1*2*3*4*5*6*7*8
==>2^7*3^2*5*7
a=3
=>b=2
a=4
=>b=2,3
a=6
=>b=2,4
a=8
=>b=3
a=12
=>b=2
7 pairs
@pavimai missed 1
==>2^7*3^2*5*7
a=3
=>b=2
a=4
=>b=2,3
a=6
=>b=2,4
a=8
=>b=3
a=12
=>b=2
7 pairs
@pavimai missed 1
@krum said:8!=1*2*3*4*5*6*7*8==>2^7*3^2*5*7a=3=>b=2a=4=>b=2,3a=6=>b=2,4a=8=>b=3so 6 pairs?
ans is 7 pairs
@pavimai said:a and b are natural numbers such that a>b>1.If 8! is divisible by a^2*b^2 ,then how much such sets (a,b) are possible??a)5b)6c)7d)8Pls share ur approach,..
8!/(a^2 b^2)
= (2^2) . (2^2) . (2^2) . 2 . 3^2 .7 .5 / (a^@ b^2)
so,
a and b not equal to 7 or 5
possible are (in the numerator)
1) 2 ^2 , 2^2
2) 2^2 , 3^2
3) 2^2 , 4^2
4) 3^2 , 2^2
5) 4^2 , 2^2
3) 4^2 , 3^2
3) 3^2 , 4^2
5) 4^2 , 2^2
3) 4^2 , 3^2
3) 3^2 , 4^2
so 7 cases
| attitude is more important than facts |
@krum said:8!=1*2*3*4*5*6*7*8==>2^7*3^2*5*7a=3=>b=2a=4=>b=2,3a=6=>b=2,4a=8=>b=3so 6 pairs?
u left out 12 ,2 pair
@pavimai said:a and b are natural numbers such that a>b>1.If 8! is divisible by a^2*b^2 ,then how much such sets (a,b) are possible??a)5b)6c)7d)8Pls share ur approach,..
8!=2^7 * 3^2 * 5 * 7
a = 3^2 => b = 2^2,2^4,2^6
a = 12^2 => b = 2^2
a = 2^4=>b = 2^2
a = 6^2=> b = 2^2, 2^4
Hence 7 pairs
a = 3^2 => b = 2^2,2^4,2^6
a = 12^2 => b = 2^2
a = 2^4=>b = 2^2
a = 6^2=> b = 2^2, 2^4
Hence 7 pairs