Paper nahin hai so solved on my hand ...started with natural solutions of a + b + c = 100 and removed cases when any of the 3 became 50 or more...but I think I did whole number instead of natural solutions so kuch galat ho gaya...didn't have patience/paper to redo hence "approx" answer bola.Edit: 1176 hi correct hai....99C2 - 3 * 50C2 = 99*49 - 3*25*49 = 24 * 49 = 1176Dubara edit: Ab paper leke aaya regardsscrabbler
Paper nahin hai so solved on my hand ...started with natural solutions of a + b + c = 100 and removed cases when any of the 3 became 50 or more...but I think I did whole number instead of natural solutions so kuch galat ho gaya...didn't have patience/paper to redo hence "approx" answer bola.Edit: 1176 hi correct hai....99C2 - 3 * 50C2 = 99*49 - 3*25*49 = 24 * 49 = 1176Dubara edit: Ab paper leke aaya regardsscrabbler
1176 is basically 49c2 where you get cases where 1 or other side is zero
isn't there a direct formula for this??n^2/48 when n is even(n+3)^2/48 when n is odd..where n is the perimeter..iss hisaab se 100^2/48 = 208 aa raha hai..
I assumed the three sides as distinct (AB, BC, AC types) ordered triplets le raha tha basically...if that is not the case then I guess the answer will be a lot smaller...and that seems a reasonable assumption as the question does not distinguish...:( So you are probably correct! regards scrabbler
A and B together can do a piece of work in 80 days. B and C can do it in 120 days. All the three started their work together on the first day, but A bunked on every odd numbered day and C bunked on every even numbered day. How long did the work last? (A) 88 days(B) 90 days(C) 96 days(D) 98 days
Guys some related questions....i cant figure out the approach to solving these questions...always get the wrong answer in these plz helpQ.1 - In how many ways can 'mn' things be distributed equally among n groups?a) mn(P)m X mn(P)n b) mn(C)m X mn(C)n c) mn!/(m!Xn!) d) mn(C)m e) None of theseQ.2 - In how many ways can a selection be made of 5 letters out of 5As, 4Bs, 3Cs, 2Ds and 1E?a) 70 b)71 c) 15(C)5 d) 15(P)5 e) None of theseQ.3 - Find the no. of ways of selecting n articles out of 3n+1, out of which n are identical.a) 2^(2n-1) b) [ (3n+1)(C)n ] / n! c) [ (3n+1)(P)n ] / n! d) 2^(2n) e) None of these
1) mn!/(m!Xn!)
2) 5 out of 5a 4b 3c 2d 1e all same -> 1 way 4 same 1 diff -> 2c1*4c1 = 8 3 same 2 diff -> 3c1*4c2 = 18 3 same 2 other same -> 3c1*3c1 = 9 2 same 3 diff -> 4c1*4c3 = 16 2 same 2 other same 1 diff -> 4c1*3c1*3c1/2 = 18 all different -> 1 total = 71 ways..
A and B together can do a piece of work in 80 days. B and C can do it in 120 days. All the three started their work together on the first day, but A bunked on every odd numbered day and C bunked on every even numbered day. How long did the work last?(A) 88 days(B) 90 days(C) 96 days(D) 98 days
A and B together can do a piece of work in 80 days. B and C can do it in 120 days. All the three started their work together on the first day, but A bunked on every odd numbered day and C bunked on every even numbered day. How long did the work last?(A) 88 days(B) 90 days(C) 96 days(D) 98 days
A and B together can do a piece of work in 80 days. B and C can do it in 120 days. All the three started their work together on the first day, but A bunked on every odd numbered day and C bunked on every even numbered day. How long did the work last?(A) 88 days(B) 90 days(C) 96 days(D) 98 days
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??a)8!/3!b)8!/6!c)5!3!d)8!/(5!3!)??
in the above question why cant the solution be 6!???like considering ABC as one entity???plz explain
let the total work be 240 units..A & B together do 3 units/dayB and C together do 2 units/dayon odd days -> b+c work = 2 unitson even days -> a+b work = 3 unitsin course of two days they complete 5 units...240/5 = 48..hence, total = 96 days...
i dont think socoz dere is no way we could find out 1/A + 1/B +1/Cso take it as A bunking on the first third fifth.... days
there actually is...
acc to the way i did the problem..
b+c=2units
a+b=3 units..
now neither wld be doing 0 units/day, so it has to be a=2 units, b=c=1 unit each...
but if they all are participating on the 1st day then the options won't suffice as the work wld then be completed in 95.4 days...which isn't in the options...so 96 hi hoga answer toh...
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??a)8!/3!b)8!/6!c)5!3!d)8!/(5!3!)??in the above question why cant the solution be 6!???like considering ABC as one entity???plz explain
8!/3!
i guess they call it the pigeon hole principle...
total arrangements/arrangements of constraints
here total arrangements = 8!
and arrangements of constraints = 3! (ABC can be arranged in 3!, but we are taking only 1 arrangement)
there actually is...acc to the way i did the problem..b+c=2unitsa+b=3 units..now neither wld be doing 0 units/day, so it has to be a=2 units, b=c=1 unit each...but if they all are participating on the 1st day then the options won't suffice as the work wld then be completed in 95.4 days...which isn't in the options...so 96 hi hoga answer toh...
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??a)8!/3!b)8!/6!c)5!3!d)8!/(5!3!)??in the above question why cant the solution be 6!???like considering ABC as one entity???plz explain
if u consider ABC as one unit it would be like ABC will always come together
but there can also be in between speakers among the three that is why u cant consider it as 6!
Given a quadratic equation ax^2+bx+c=0, and X1, X2 are two distinct roots and |X1|=|X2|, then which of the following is always correct a) |b||c| d) |a|
