Try it

It is not at all hard , ( not only from the view point of sirs ) , so try this ----

1 > Let a , b , c, d , e …… be the positive divisors of “ n “ except n and 1 . Prove that ---------

( 1 / a ) + ( 1 / b ) + ( 1 / c ) + ( 1 / d ) + …… = p (n ) / n ,

Where p ( n ) = a + b + c + d + e ………..

11 Answers

1
akari ·

well i have seen this b4 so i will not post :
hint to begginers :think on multiplicative inverse

1
Maths Musing ·

Yeah , dukhia .. , you are absolutely correct about the hint ..
Try another one ,

Find the number of words that can be made from the word “ TRIANGLE “ such that the relative order of vowels do not change .

1
Maths Musing ·

Ans is 6720 , and dukhi.. you have considered only i and e whereas a is also there in the word , which is why your ans. is very small .
in your ans , several cases as A_I_ _ _ _E have been considered which should be omitted

1
Maths Musing ·

The least I can say , the solution too was anything beyond my imagination , but still I managed to do it :):)

1
Maths Musing ·

small hint -- don't go in taking cases at all , just try to substitute something instead of something

1
Maths Musing ·

I think your process is correct but there must be a calculation mistake.

1
Maths Musing ·

Isn't the case _ _ _ I A E _ _ being calculated thrice ?

62
Lokesh Verma ·

There is a much simpler ways guys...

n! / 3!

thnk it out [1]

1
Maths Musing ·

no need to say " Absolutely correct sir " , I wonder from where these brainwaves come !!!!!!!!

but then of course , that is why you are nishant sir

1
Maths Musing ·

Yeah , I did it by that process only.

1
Maths Musing ·

Replace the " I , A and E " by 3 I's .
so you get the following letteres to form words --- T , R , N , G , L , I , I , I
observe that anyhow you arrange them , you are gonna get the relative order of vowels same , coz all the vowels are same only !!!!
so no.of words = no. of words that can be formed anyhow = 8! / 3!
similarly you can get this result for n lettered words also.

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