yuktimittal's posts

yuktimittal
replied to doubts in permutation & combination

the question asked in this thread asks about angle abc being acute whereas in the link u suggested the question deals with triangle abc being acute..

ps- thanks, coz the link u hv given is vry useful.

ps- thanks, coz the link u hv given is vry useful.

yuktimittal
replied to Quant by Arun Sharma

no. of ways 4 nt selecting 8, 9 and 10 =7 /10.

as we have to select 6 tickets thus no. of ways = 7^6/10^6

bt in the selected 6 tickets the highest no can also be 6 (1,2,3,4,5,6 numbered ticket) thus we subtract the cases of highest no.on any of the 6 tickets being 6 which is not selecting 7...

as we have to select 6 tickets thus no. of ways = 7^6/10^6

bt in the selected 6 tickets the highest no can also be 6 (1,2,3,4,5,6 numbered ticket) thus we subtract the cases of highest no.on any of the 6 tickets being 6 which is not selecting 7...

yuktimittal
replied to doubts in permutation & combination

2nd question:total cases 4 d ranks will b 50 *50

n d cases where d differences b/w thr ranks is not more than 10 will be 940( to calculate 940 consider ranks indivisually for eg. if aryan's rank is 1 nandini can hv 11 ranks;aryans rank is 2 den nandini can hv 12 ranks dis goes on increasin ...

n d cases where d differences b/w thr ranks is not more than 10 will be 940( to calculate 940 consider ranks indivisually for eg. if aryan's rank is 1 nandini can hv 11 ranks;aryans rank is 2 den nandini can hv 12 ranks dis goes on increasin ...

yuktimittal
replied to doubts in permutation & combination

ans 2. 9/25

ans 3. 3/4

r the anwers right?

ans 3. 3/4

r the anwers right?

yuktimittal
replied to doubts in permutation & combination

ans. 3. 6/55

total positons that r and e can take when placed randomly= 11*10

no. of positions that r and e can take when there are exctly 4 letters between them =12

so 12/(11*10)= 6/55

total positons that r and e can take when placed randomly= 11*10

no. of positions that r and e can take when there are exctly 4 letters between them =12

so 12/(11*10)= 6/55

yuktimittal
replied to doubts in permutation & combination

thnx dint see dat in a hurry plus i dint do d last step

yuktimittal
replied to doubts in permutation & combination

hey if the question was that any one of the boxes contains 2 balls,then 4*45*3^8 would b the right ans..

yuktimittal
replied to doubts in permutation & combination

d 2nd question can also be done by cyclicity of 7.. dats is

7^1=7

7^2=49

7^3=343

7^4=2401

so we hv 7,9,3,1 which will repeat.. so till 100 thr will b 25 such sets n for m=1 n=3(7+343=350).. thr will b a total of 25 values for n for 1 value of m and if m=2 den n=4(49+2401) dis wil...

7^1=7

7^2=49

7^3=343

7^4=2401

so we hv 7,9,3,1 which will repeat.. so till 100 thr will b 25 such sets n for m=1 n=3(7+343=350).. thr will b a total of 25 values for n for 1 value of m and if m=2 den n=4(49+2401) dis wil...