@scrabbler said:21 ways?regardsscrabbler
Bhai approach batao...
@bs0409 said:The infinite sum1 + 4/7 + 9/7^2 + 16/7^3 + 25/7^4 +.....equalsa.27/14b.21/13c.49/27d.256/147
Sometimes it is difficult to see the mathematical pattern involved. However, if you have the time (~2-3 minutes!), and are fast with calculations, you can still get your 3 marks with a little luck. Or at least, eliminate some options...
Let's add the first few terms. 1 + 4/7 + 9/49 + 16/343 + 25/2401...= 1 + 0.57 + 0.18 + 0.048 + 0.01 + = 1.808 + very small terms (note that each further term is less than 1/3 of the previous)
Now the 2nd and 4th options are already below this and can be eliminated, 1st is 1.92 which is never going to be reached, and 3rd is 1.81....reasonable risk.
Obviously if you can get a guaranteed answer it is nicer 😃 but in case you can't see a way to start it mathematically, this could help...
regards
scrabbler
Let's add the first few terms. 1 + 4/7 + 9/49 + 16/343 + 25/2401...= 1 + 0.57 + 0.18 + 0.048 + 0.01 + = 1.808 + very small terms (note that each further term is less than 1/3 of the previous)
Now the 2nd and 4th options are already below this and can be eliminated, 1st is 1.92 which is never going to be reached, and 3rd is 1.81....reasonable risk.
Obviously if you can get a guaranteed answer it is nicer 😃 but in case you can't see a way to start it mathematically, this could help...
regards
scrabbler
@bs0409 said:OA-49/27in how many ways 7^13 can be written as product of three natural nos
15?? :neutral:
@saurav205 said:Bhai approach batao...
bhai dekho... 15c2 is 105..
but in 105 cases......cases like (0,13,0); (11,1,1) etc are counted 3 times (3!/2! times) these are 7*3 = 21 cases...hence 105-21 = 84 cases are left...
but these 84 cases are counted 6 times (3! times)...so 84/6 = 14 cases (unordered)...
so total = 14+7 = 21 unordered...
but iska ek direct bhi hota hai ek traika...was looking for that...i hope @scrabbler can help here...
@Logrhythm said:21??7^(a+b+c) where a+b+c = 13..i am getting 21 such cases after manual counting...can someone let me know how to work on calculating the un-ordered pairs once we arrive at 15c2??
I don't think that is possible. Maine bhi manually likhe...working from here to get to 15C2 is still feasible, but the reverse side, doubtful...
@Logrhythm said: aapne toh manually nahi count kara??
Manually only....
regards
scrabbler
@Logrhythm said: aapne toh manually nahi count kara??
Manually only....
regards
scrabbler
@scrabbler ye wala dekho fir zara...
In how many ways 2310 be written as product of 3 different factors??
@amanaspirant said:Hi all,Please provide the solution of below question.The sum of three consecutive even integersis 132. Find the difference between3 times the least and half of the greatest.1) 44 2) 893) 144 4) 103
x+x+2+x+4= 132.. solve for x and then whatever *3 and half etc that are required.. ans is 103.. :)
@Logrhythm said:bhai dekho... 15c2 is 105..but in 105 cases......cases like (0,13,0); (11,1,1) etc are counted 3 times (3!/2! times) these are 7*3 = 21 cases...hence 105-21 = 84 cases are left... but these 84 cases are counted 6 times (3! times)...so 84/6 = 14 cases (unordered)...so total = 14+7 = 21 unordered...but iska ek direct bhi hota hai ek traika...was looking for that...i hope @scrabbler can help here...
Haan yar...
Bus main hoon..so manually counting is difficult....
Koi aur approach....
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
10...
@Logrhythm said:bhai dekho... 15c2 is 105..but in 105 cases......cases like (0,13,0); (11,1,1) etc are counted 3 times (3!/2! times) these are 7*3 = 21 cases...hence 105-21 = 84 cases are left...but these 84 cases are counted 6 times (3! times)...so 84/6 = 14 cases (unordered)...so total = 14+7 = 21 unordered...but iska ek direct bhi hota hai ek traika...was looking for that...i hope @scrabbler can help here...
Let the numbers be 7^a,7^b,7^c
so, 7^(a+b+c)=7^13
a+b+c=13
So,15C2=105 solns
But we want unordered solutions
When two of a,b,c are same:2a+c=13 [since b=a]
which has 7 solutions.
So 3C2*7=21 cases for 2 of them to be same
All three same i.e. a=b=c: not possible
So Unordered pairs= (105-21)/3! + 21/3=14+7=21 ways
@bs0409 said:Let the numbers be 7^a,7^b,7^cso, 7^(a+b+c)=7^13a+b+c=13So,15C2=105 solnsBut we want unordered solutionsWhen two of a,b,c are same:2a+c=13 [since b=a]which has 7 solutions.So 3C2*7=21 cases for 2 of them to be sameAll three same i.e. a=b=c: not possibleSo Unordered pairs= (105-21)/3! + 21/3=14+7=21 ways
haan essentially this is what i did in the post above....sahi hai.. thanks :)
@bs0409 said:Let the numbers be 7^a,7^b,7^cso, 7^(a+b+c)=7^13a+b+c=13So,15C2=105 solnsBut we want unordered solutionsWhen two of a,b,c are same:2a+c=13 [since b=a]which has 7 solutions.So 3C2*7=21 cases for 2 of them to be sameAll three same i.e. a=b=c: not possibleSo Unordered pairs= (105-21)/3! + 21/3=14+7=21 ways
bhai yeh kaise aaya?
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
Eska oa kya hai???
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
Is it 41 by any chance?
2310 = 2*3*5*7*11
No of ways when all factors are diff = (3^5-3)/3!
When 2 are same, i.e, 1 = 1 way (1,1,2310)
Total = 41 ways ?? ..
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
Ummm....41? aila confusing hai yeh waala!
regards
scrabbler
regards
scrabbler
@Logrhythm said:bhai dekho... 15c2 is 105..but in 105 cases......cases like (0,13,0); (11,1,1) etc are counted 3 times (3!/2! times) these are 7*3 = 21 cases...hence 105-21 = 84 cases are left... but these 84 cases are counted 6 times (3! times)...so 84/6 = 14 cases (unordered)...so total = 14+7 = 21 unordered...but iska ek direct bhi hota hai ek traika...was looking for that...i hope @scrabbler can help here...
sir nai samajh aaya.. ek bar phir se samjhao.. aapne manually cases nikale h kya??? if yess then tell me.. 
@scrabbler said:Ummm....41? aila confusing hai yeh waala!regardsscrabbler
Apka confusion bhi humari surity se sahi hota h :P
@ScareCrow28 said:Is it 41 by any chance?2310 = 2*3*5*7*11No of ways when all factors are diff = (3^5-3)/3!When 2 are same, i.e, 1 = 1 way (1,1,2310)Total = 41 ways ?? ..
Nice approach! Mine was much longer...
regards
scrabbler
regards
scrabbler
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
175?