Given ans is ryt...req no is Hcf of diff b/w ny two nos and not 3 as u did...:-)
No...it
ans is 123.
req no. is the HCF of diff b/w all the 3 nos.
ans is 123.
req no. is the HCF of diff b/w all the 3 nos.
ans is 123.
req no. is the HCF of diff b/w all the 3 nos.
going by the TIME it shud be HCF of diff b/w ny two nos.....cud u plz tell share source of info on dis...
how many 4 digit even numbers can be formed using digits 1,2,3,4,5,6(repetition is allowed)?
1)648
2)180
3)1296
4)540
5)600
My take 1)648.
unit digit will be even so only three numbers for unit digit position(2,4,6).
For the remaining positions,we have 6 numbers will repition allowed.
so it will be 6 X 6 X 6 X 3 = 648 . . .
1)There are 8 orators A,B,C,D,E,F,G,H.In how many ways can arrangements be made so that A always comes before B and B always comes before C.
1)8!/3!
2)8!/6!
3)5!*3!
4)8!/(5!*3!)
2)Seven objects must be divided among 3 people.In how many ways can this be done if one or two of them can get no objects.
1)15120 2)2187 3)3003 4)792
3)In how many ways can 10 identical marbles be distributed among 6 children so that each child gets at least 1 marble?
1)15C5 2)15C9 3)10C5 4)9C5
4)The crew of an 8 member rowing team is to be chosen from 12 men,of which 3 must row only on one side and 2 must row on the other side only.find the number of ways of arranging the crew with 4 members on each side.
1)40320 2)30240 3)60480 4)10080 5)None of These
1)There are 8 orators A,B,C,D,E,F,G,H.In how many ways can arrangements be made so that A always comes before B and B always comes before C.
1)8!/3!
2)8!/6!
3)5!*3!
4)8!/(5!*3!)
We just need to select 5 places for D, E, F, G and H. That is because we have a set pattern for A, B and C. i.e. there's just 1 way to do so. now arrangement would be made in 8P5 ways. i.e. = 8!/3!. Hence option 1.
4)The crew of an 8 member rowing team is to be chosen from 12 men,of which 3 must row only on one side and 2 must row on the other side only.find the number of ways of arranging the crew with 4 members on each side.
1)40320 2)30240 3)60480 4)10080 5)None of These
In the above question we have 5 positions taken up so we are left with 7 people and only 3 need to be selected. This can be done in 7C3 ways. i.e. = 7!/(3!4!). Now rowers on each side can be placed in 4! ways and there are 2 sides to select for each group of 4 rowers. So the arrangement becomes.... (7!/(3!4!))*4!*4!*2! which gives the answer as 40320 hence option 1.
2)Seven objects must be divided among 3 people.In how many ways can this be done if one or two of them can get no objects.
1)15120 2)2187 3)3003 4)792
First object can be given to any of the 3 and similarly for other six objects
=> 3^7 = 2187 ways
3)In how many ways can 10 identical marbles be distributed among 6 children so that each child gets at least 1 marble?
1)15C5 2)15C9 3)10C5 4)9C5
a is the number of marbles given to first child
b is the number of marbles given to second child
.
.
f is the number of marbles given to sixth child
=> a + b + c + d + e + f = 10
But all of them are atleast 1
=> a = a' + 1, b = b' + 1, ....
=> a' + b + c' + d' + e' + f' = 4
=> C(4 + 6 - 1, 4) = C(9, 5) ways
4)The crew of an 8 member rowing team is to be chosen from 12 men,of which 3 must row only on one side and 2 must row on the other side only.find the number of ways of arranging the crew with 4 members on each side.
1)40320 2)30240 3)60480 4)10080 5)None of These
Choose the remaining 3 in C(7, 3) ways. Now, any two the three will go the side where there are just two in C(3, 2) ways
Now, we can also arrange the 4 persons on both the sides in 4!*4! ways
=> Total = C(7, 3)*3*4!*4! = 60480
But all of them are atleast 1
=> a = a' + 1, b = b' + 1, ....
=> a' + b + c' + d' + e' + f' = 4
=> C(4 + 6 - 1, 4) = C(9, 5) ways
I understand that the formula for distributing R things among N people when some can get none is C((R+N-1), (N-1)) but could you explain the logic behind this?
UtsavGambhir SaysI understand that the formula for distributing R things among N people when some can get none is C((R+N-1), (N-1)) but could you explain the logic behind this?
Suppose we have to distribute n objects among r persons, then
Say 1st person will get a1 objects
2nd person will get a2 objects
.
.
rth person will get ar objects
=> a1 + a2 + a3 + ... + ar = n
So, basically we have to find the non-negative solution for this eq.
Sin ak can vary from 0 to n for all k from 1 to r.
Number of non-negative integral solutions
= Coeff of x^n in (1 + x + ... + x^n)^r
= Coeff of x^n in (1 - x^(n + 1))^r * (1 - x)^(-r)
= Coeff of x^n in (1 + C(r, 1)x + C(r + 1, 2)x^2 + .... C(r + n - 1, n)x^n + ...)
= C(n + r - 1, n) or C(n + r - 1, r - 1)
1) A train is running between Patna to Howrah.seven people enter the train somewhere between Patna and Howrah.It is given that there are 9 stops between Patna and Howrah.In how many ways can tickets be purchased if no
restriction is there with respect to the number of tickets at any station?
1)45C7
2)63C7
3)56C7
4)52C7
1)There are 8 orators A,B,C,D,E,F,G,H.In how many ways can arrangements be made so that A always comes before B and B always comes before C.
1)8!/3!
2)8!/6!
3)5!*3!
4)8!/(5!*3!)
For such Questions -
The Number of ways of arranging all 8 orators = 8!
Because the relative position of ABC is fixed - i.e. A before B and B before C, there fore - out of 6 possible arrangements of A,B and C - only one arrangement will be such that A is before B and B is before C - therefore in every 6 arrangements of A,B and C only 1 will obey the condition given -
therefore from 8 ! arrangments of all orators containing A, B and C
only 8! /6 will obey the condition - Ans = 8! /6 = 8! /3!
if a year is selected at random what is the probability that is has 53 mondays?
hey puys, please lemme know which book is better for me?arun sharma or quantum cat arihant,,, as i was enrolled in TIME earlier n cat 2011 will be my second attempt.I want quality questions of a good level to be there. I knw arun sharma doesnt contain answers..n arihant does. but apart from explaining concepts very nicely and elaborating answers, does arihant have a gud high level quants questions???? Please guys help me out here... thnx 😃
zenith12 Sayshey puys, please lemme know which book is better for me?arun sharma or quantum cat arihant,,, as i was enrolled in TIME earlier n cat 2011 will be my second attempt.I want quality questions of a good level to be there. I knw arun sharma doesnt contain answers..n arihant does. but apart from explaining concepts very nicely and elaborating answers, does arihant have a gud high level quants questions???? Please guys help me out here... thnx :)
take any book d sole aim is to practice more n more watever d material is I did arun sharma till april and now I m not doing it m referring time material oly for every topic............so u can take anything but jus
practice practice and practice
take any book d sole aim is to practice more n more watever d material is I did arun sharma till april and now I m not doing it m referring time material oly for every topic............so u can take anything but jus
practice practice and practice
hey thnx...hw did u find the questn level..n as the answers are not there..did u face ne difficulty?? hv practiced the time maerial earlier..just wanna a gud buk nw.. pls lemme knw..
zenith12 Sayshey puys, please lemme know which book is better for me?arun sharma or quantum cat arihant,,, as i was enrolled in TIME earlier n cat 2011 will be my second attempt.I want quality questions of a good level to be there. I knw arun sharma doesnt contain answers..n arihant does. but apart from explaining concepts very nicely and elaborating answers, does arihant have a gud high level quants questions???? Please guys help me out here... thnx :)
For high quality questions an explaination i believe pagal guy is the best..so stick with the site..
there are four machines and it is known that exactly two of them are faulty.they are tested one by one in a random order till both the faulty machines are identified.then the probability that exactly three tests will be required to identify the faulty machines will be
1)1/2
2)1
3)1/3
4)2/3
vegeta Saysif a year is selected at random what is the probability that is has 53 mondays?
Probability of selecting a non leap year = 3/4
and when the year has 365 days then there would be 1 day left after completion of 52 weeks. So, probability of that day being a monday would be 1/7.
Probability of selecting a leap year would be 1/4
and when the year has 366 days days then there would be 2 days left after completion of 52 weeks. So, probability of that day being a monday would be 2/7. As monday can appear in 2 combinations i.e. Sunday, Monday and Monday, Tuesday.
So, overall probability = (3/4*1/7)+(1/4*2/7) = 5/28
there are four machines and it is known that exactly two of them are faulty.they are tested one by one in a random order till both the faulty machines are identified.then the probability that exactly three tests will be required to identify the faulty machines will be
1)1/2
2)1
3)1/3
4)2/3
3 tests would be required in 2 cases.
When the order of selection is
1) faulty, non-faulty, faulty
2) non-faulty, faulty, faulty
Probability of 1) happening is 2/4*2/3*1/2
and the probability of 2 happening is also same i.e. 2/4*2/3*1/2
Hence overall probability = 2*(2/4*2/3*1/2) = 1/3
Can anybody give me the answer to this question?
There is a 200 miles long tunnel. one train enters the tunnel at a speed of 200mph while the other trains enter the tunnel in the opposite direction at a speed of 1000 mph. A bee travels at a speed of 1500 mph enters the tunnel goes to and back until it reaches the train. What is the distance covered by the bee when the two train collides (the bee survives)
Thanks