if `6n` tickets numbered 0,1,2,....... 6n-1 are placed in a bag and three are drawn out , show that the chance that the sum of the numbers on then is equal to 6n is `(3n)/((6n-1)(6n-2))`

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

A bag contains n coupons marked 1,2,3 , …. , n .If two coupons are drawn , then the chance that the difference of the coupons exceeds m ( les than n - 1) is :

State the number of significant 6.032 N m^(-2)

Statement-1: Out of 21 tickets with number 1 to 21, 3 tickets are drawn at random, the chance that the numbers on them are in AP is (10)/(133) . Statement-2: Out of (2n+1) tickets consecutively numbered three are drawn at ranodm, the chance that the number on them are in AP is (4n-10)/ (4n^(2)-1) .

Let N underseta = (an : n in N) , then N underset6 in N underset8 is equal to

Use proof by contradiction to show that there is no value of n for which 6n ends with the digit zero.

If X={4^n -3n-1 , n in N} and Y={9(n-1), n in N} , where N is the set of natural numbers, then X uu Y is equal to :

Check whether 6^n can end with the digit 0 for any natural number n.

Let n(A)=3 and n(B)=6 and A ulsub B. Then number of elements is A nn B is :

The sum of n terms of the series 1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+... is