Now try to solve this:(it's interesting)
In how many ways one or more of 5 different letters be posted to 4 different letter boxes?
I will post the answer later..
Good luck!!
4^5 ways is the answer
4^5 ways is the answer
each letter can be posted in 5 ways...i.e in 4 boxes and also 1 way not to post it,
hence total ways is 5^5 ways but in this there is 1 way in which no letter is posted so v need to subtract tht...hence the required ans is 5^5 -1
I didn't quite understand this. Could you please explain it?
Hi guys, help me out with this prob.
Without stoppage, train travels a distance wt. an avg speed 60kmph, with stoppage, with an avg speed of 40 kmph. On an avg, how many minutes per hour does train stop?
a. 20 min/h
b. 15 min/h
c. 10 min/h
d. 12 min/h
Hi guys, help me out with this prob.
Without stoppage, train travels a distance wt. an avg speed 60kmph, with stoppage, with an avg speed of 40 kmph. On an avg, how many minutes per hour does train stop?
a. 20 min/h
b. 15 min/h
c. 10 min/h
d. 12 min/h
Should be 20 min/h
Hi guys, help me out with this prob.
Without stoppage, train travels a distance wt. an avg speed 60kmph, with stoppage, with an avg speed of 40 kmph. On an avg, how many minutes per hour does train stop?
a. 20 min/h
b. 15 min/h
c. 10 min/h
d. 12 min/h
avinav2712 SaysShould be 20 min/h
Yes,20 mins is the answer (pretty sure).
Although this can be done from bit of mental calculation as well , I am providing bit detailed explanation :
Let the distance be 60 kms.So time taken in latter case = 1.5 hrs
which is 30 mins extra than usual...30 mins extra in 90 mins...so,by unitary method...20 mins extra in 60 mins..20m / hr..
hope this helps
Try solving this questions with the data - 75 kms per hour & 60 kms per hour respectively...find stoppage time per hr...answer is 12 mins..m sure u will handle it now :)
best of luck!!
inava SaysI didn't quite understand this. Could you please explain it?
each letter can be posted in 5 ways....that is in 4 boxes and 1 not posting it at all...so using the PNM rule all the 5 letters can be posted in 5*5*5*5*5 = 5^5 ways. here as they are say atleast 1 letter shud be posted we subtract 1 from the above which is the case when no letter is posted. hence the total ways will be 5^5-1
Quoting the Questions from LOD(III) - Permutations and Combinations
1. The number of ways in which four particular persons A,B,C,D and six more persons can stand in a queue so that A always stands before B, B always before C and C always before D is?
1) let xa be the no of guys standing before A >=0
Let xb be the no. of guys standing between A and B >=0
Let xc be the no. of guys standing between B and C >=0
Let xd be the no. of guys standing between C and D >=0
Let x be the no. of guys standing after D >=0
now Xa + Xb+ Xc+ Xd + X = 6
hence the total arrangement possible will be 6 + 5 - 1 C 5 - 1 = 10C4 = 210..
Is this the ans?
1) let xa be the no of guys standing before A >=0
Let xb be the no. of guys standing between A and B >=0
Let xc be the no. of guys standing between B and C >=0
Let xd be the no. of guys standing between C and D >=0
Let x be the no. of guys standing after D >=0
now Xa + Xb+ Xc+ Xd + X = 6
hence the total arrangement possible will be 6 + 5 - 1 C 5 - 1 = 10C4 = 210..
Is this the ans?
Let us select 4 places out of the ten at which abcd will stand which will be given by 10C4 Now for this selection it will always follow an order a-b-c-d irrespective of the number of positions in between Now we can arrange other 6 guys in factorial 6 ways and hence answer should be 6! * 10 C4 Is the answer correct ď Š
abhay15783 SaysLet us select 4 places out of the ten at which abcd will stand which will be given by 10C4 Now for this selection it will always follow an order a-b-c-d irrespective of the number of positions in between Now we can arrange other 6 guys in factorial 6 ways and hence answer should be 6! * 10 C4 Is the answer correct ď Š
hey u r rt dude...aah:w00t: y am i making such silly mistakes
forgot to arrange them...thnz for the help dude
2. The number of circles that can be drawn out of 10 points of which 7 are collinear is ?
The number of ways will be no points are collinear+selecting 2 points collinear+selecting 1 collinear point =3C3+7C2*3+7*3 =1+63+21 =85 circles are possible correct me if i am wrong
1) let xa be the no of guys standing before A >=0
Let xb be the no. of guys standing between A and B >=0
Let xc be the no. of guys standing between B and C >=0
Let xd be the no. of guys standing between C and D >=0
Let x be the no. of guys standing after D >=0
now Xa + Xb+ Xc+ Xd + X = 6
hence the total arrangement possible will be 6 + 5 - 1 C 5 - 1 = 10C4 = 210..
Is this the ans?
Can u elaborate on Xa + Xb+ Xc+ Xd + X = 6
why total = 6 ?
Can u elaborate on Xa + Xb+ Xc+ Xd + X = 6
why total = 6 ?
those 6 guys other than A B C and D who can be placed anywhere between A B C and D. I hope u got it:)
shanks4mba Saysthose 6 guys other than A B C and D who can be placed anywhere between A B C and D. I hope u got it:)
4. A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?
in the same way
1,2,3,4,5,6 be a,b,c,d,e,f respectively
a>0,b,c,d,e,f>=0
(a-1)+b+c+d+e+f = 5
a+b+c+d+e+f = 6
6+6-1c5 = 11c5
correct me if i am wrong
4. A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?
I think the answer will be 6c2* 4! can we have the official answer
4. A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?
in the same way
1,2,3,4,5,6 be a,b,c,d,e,f respectively
a>0,b,c,d,e,f>=0
(a-1)+b+c+d+e+f = 5
a+b+c+d+e+f = 6
6+6-1c5 = 11c5
correct me if i am wrong
no...not this way...let a be the nos. coming before 1 and b be nos. btw 1 and 6 and c be nos. after 6. I will leave the rest for u to solve:) revert back in case of doubt...
Following doubts:
1. Que. 1.......the ans will be 9:16 or 9:25... m getting 9:25
2. Que. 7.......detail solution required
3. Que. 19..... detail solution required
time speed and distance of arun sharma lod1 question no.s 19
plzzzzzzzz ans dis 1 plzzzz
time speed and distance of arun sharma lod1 question no.s 19
plzzzzzzzz ans dis 1 plzzzz
Kindly always post the question over here and dont ask from question numbers you see most of guys are not sitting with arun sharma ( in fact most will be in office just like me 😃 )
Guys I need to ask you, how r u going through this book? In a traditional way like doing basics then lod1,2,3 or you have some other strategy?
time speed and distance of arun sharma lod1 question no.s 19
plzzzzzzzz ans dis 1 plzzzz
let Manoj needs 'k' days to a peice of work.
so, Anjay will require k/2 days and Vijay will need k/3 days to do the same.
M's 1 day work = 1/k
A's =2/k
v's =3/k
M+A+V combined 1 day work = 1/k + 2/k +3/k
=6/k
now give they finish the work in 1 day , so the work they have done together in one day is the total work
total work=6/k
now M's one day work = 1/k
no. of days required = total work/ one days work
" " =6