if number of function defined from f={1,2,3,4,5,6}→{7,8,9,10} such tht the sum of f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=odd number is 2n .then find the value of n???
106sum of six nos = odd
=> 5 even + 1 odd
3 even + 3 odd
1 even + 5 odd
now can you find the no. of cases possible ?
106what i meant in my prev. post is that in case III, five elements will have an odd image and one will have an even image.
in case II, three elements will have an odd image and threee will have an even image.
in case I, five elements will have an even image and one will have an odd image.
14 ways of choosing f(1)
4 ways of choosing f(2)
.
.
4 ways of choosing f(5)
now, for every set of {f(1),f(2)..f(5)}
f(1)+f(2)..f(5) can either be odd or even,
if it is even, for the final sum (ie f(1)+..f(6)) to be odd, we have to choose from {7,9}.
if it is odd, for the final sum(ief(1)+..f(6)) to be odd, we have to choose from {8,10}
notice in either case we have 2 final outcomes, hence total ways= 4*4..(5 times)*2
=45*2=211
another possible way of doing this is by intuition, out of all mappings, sum can be either odd or even, which has equal chances. hence for each event, the probability=1/2, or ways for each= 2^12 *1/2
1@goedo sir , the second solution is nicei had the same thing in my mind