1yeah you are right it is "the set of" !!
let S be the set of 6 -digit numbers a1 a2 a3 a4 a5 a6 (all digits are distinct) where a1≥ a2≥ a3≥ a4≤ a5≤ a6.then n(S) is equal to NOTE :PLEASE IGNORE THE EQUALITY AND CONSIDER ONLY THE INEQUALITY RELATIONS .(i could not find the signs of inequalities without equalities) a.210 b.2100 c.4200 d.420
-
UP 0 DOWN 0 0 4
4 Answers
62you have to find
a4=k +1 (let)
a3=a4+x+1
a2=a3+y+1
a1=a2+z+1
a5=a4+w+1
a6=a5+v+1
x,y,z,w,v are all >1
a1=k+x+y+z+4
a6=k+w+v+3
now we also know that
a1<=9
a6<=9
k+x+y+z+4<=9
k+w+v+3<=9
k+x+y+z<=5
k+w+v<=6
now take k=0,1, so on till 5
and for each case find the number of solutions and add
PS: This method may sound very vague and useless but try to understand. It will help in solving all problems of this type..
62The above was the most stupid solution i ever gave... close to that...
no of ways to chose 6 out of 10 numbers 0 to 9 = 10C6
See a4 is clearly the smallest of all the numbers..
There is only 1 way to pick a4
now you have to pick a1,a2,a3 just pick 3 from the remaining 5
5C3=10 ways
Now for each choice, there is only 1 way in which a1>a2>a3. Hence the number of ways to chose them is 10
Thsu the total no of ways is 2100 [1]