IIT JEE , AIEEE Forum - Mathematics , Algebra - permutation and combinations again!!!!!!!!

abhijeet_0201

hey ppl plz try this tough prob(for me at least) on p&c

the no of ways of choosing 2 integers frm {1,2,3,............100}

such that their difference is atmost 10.

(a)900 (b)935 (c)945 (d)3005

plz dont ask me the ans or say that the ques is inc or wrong.the ques is as it is and its source is trustworthy(bt test)

plz tell waht u think shud be the ans .plz reply soon.

waiting..........................

abhijeet_0201

come on..................................

KAB

The answer is c)945

Let a and b be two numbers.

The condition is |a-b|<=10

When a=1 b can be 2,3,4,5,6,7,8,9,10,11 (I've neglected a-b=0 since none of the option matches the answer.)

So there are 10 ways.

Same thing happens till a=90.

So no. of ways is 90*10=900

Now when a=91 b can be 92,93,94,95,96,97,98,99,100

ie 9 ways

When a=92 there are 8 ways.

So on when a=100 there is only 1 way.

So total no. of ways=900+9+8+....+1=

=900+9*10/2

=945

abhijeet_0201

phenomenal!!!!!!!!!!!!!!!!!!!!!!
awesome!!!!!!!!!!!!!!!!!!!!!!!!!!

joyfrancis

First case (difference is 1)

numbers that can be formed are

(1,2) (2,3) (3,4)...........(99,100) total ways=99

Second case(difference=2)

numbers are

(1,3) (2,4) (3,5) (4,6)........(98,100) total ways=98

Third case(difference 3)

(1,4) (2,5) (3,6).........(97,100) total ways = 97

You can see that the number of ways is decreasing by1,so if we are asked to go upto 10 the number of ways would be

99+98+97+96......10terms

or

90+91+92.........99

By sum of AP

S=945

magiclko

Adarsh and joyfrancis both has done a gr8 job!!!