Maths

[anon_econ]

In how many ways can 6 distinct integers be selected from the set {1,2,3...49} such that no 2 consecutive integers are selected?

[s]

hi Vasudha is the answer 7C0* 44C38 = 44C6  ?

[anon_econ]

Shreya I don't know the correct answer..but how have u done this??

[s]

out of 49 we have to select 6 integers.

so let 'a' integers be there before the first chosen integer, 'b' integers between 1st and 2nd chosen integer, 'c' integers between 2nd and 3rd chosen integers, 'd' integers between 3rd and 4th, 'e' between 4th and 5th, 'f' between 5th and 6th, 'g' after 6th integer

so it looks like this:

....a integers....O....b integers....O..c integers...O...d integers...O...e integers...O...f inetgers....O...g integers...

O is a chosen integer,

now a+b+c+d+e+f+g=49-6=43 the remaining integers after choosing 6 integers must add up to 43..

now apply multinomial theorem..
a can take values 0,1,2,3,...,38  cant be more than 38 (check for yourself)
b can take values 1,2,3,.....,39
c can take values 1,2,3.....,39
d same as above
e same as above
f same as above
g can take 0,1,2,....,38

now find coefficient of x^43 in ( x^0 + x^1+...+x^38)(x^1 + x ^2 +...x^39)...(x^0+x^1+...x^38)

try it yourself and see what answer you get

you can find sums like these in any maths mcq book,its a very effective method to solve sums like these

[anon_econ]

Thanks a ton shreya :) :)
I had never heard of the multinomial theorem. I had to look it up right now to understand the last step. And since such questions can be found in 'any maths mcq book' i clearly need to work hard :P

[anon_econ]

But how do we find the coefficient of x^43??

[s]

hey, basically if you simplify you get..

(x^0+x^1+...+x^38)^2(x^1+x^2+...x^39)^5
=(x^0+x^1+...+x^38)^2 * x^5*(1+x^1+...+x^38)^5
=x^5*(x^0+x^1+...+x^38)^7

 now x^0+x^1+...+x^38 is a GP so find its sum...i got something like
x^5* (1-x^39/1-x)^7
= x^5 * (1-x^39)^7* (1-x)^-7
 
now expand (1-x)^-7 as 1+7C1*x+ 8C2* x^2 +......to infinity (look up formula for this in Binomial theorem chapter)

next alongside expand (1-x^39)^7

now search for the coefficient of x^43 in the simplified expression..
its a little cumbersome for the first time but its really effective...if you have Tata Mcgraw Hill's Maths mcq(for IIT) look up the permutations chapter for sums like these

[anon_econ]

Thank u shreya :)