The number of ways in which we can distribute `m n` students equally among `m` sections is given by a. `((m n !))/(n !)` b. `((m n)!)/((n !)^m)` c. `((m n)!)/(m ! n !)` d. `(m n)^m`

Find the number of ways in which n different prizes can be distributed among m(< n) persons if each is entitled to receive at most n-1 prizes.

2m white counters and 2n red counters are arranged in a straight line with (m+n) counters on each side of central mark. The number of ways of arranging the counters, so that the arrangements are symmetrical with respect to the central mark is (A) .^(m+n)C_m (B) .^(2m+2n)C_(2m) (C) 1/2 ((m+n)!)/(m! n!) (D) None of these

Prove : int sin mx sin n x dx[ m^(2) != n^(2)] , = 1/2 [ (sin(m-n)x)/(m-n) - (sin (m+n)x)/(m+n) ] + c

A function f is defined by f(x)=|x|^m|x-1|^nAAx in Rdot The local maximum value of the function is (m ,n in N) (a) 1 (b) m^n.n^m (c) (m^m n^n)/((m+n)^(m+n)) (d) ((m n)^(m n))/((m+n)^(m+n))

If the sum of m terms of an A.P. is the same as the sum of its n terms, then the sum of its (m+n) terms is (a). m n (b). -m n (c). 1//m n (d). 0

Straight lines are drawn by joining m points on a straight line of n points on another line. Then excluding the given points, the number of point of intersections of the lines drawn is (no tow lines drawn are parallel and no these lines are concurrent). a. 4m n(m-1)(n-1) b. 1/2m n(m-1)(n-1) c. 1/2m^2n^2 d. 1/4m^2n^2

The number of ordered pairs of positive integers (m,n) satisfying m le 2n le 60 , n le 2m le 60 is

The number of ways of arranging m positive and n(lt m+1) negative signs in a row so that no two are together is a.= ^m+1p_n b.= ^n+1p_m = c.= ^m+1c_n ="" d.= ^n+1c_m

The number of ordered pairs (m,n) where m , n in {1,2,3,…,50} , such that 6^(m)+9^(n) is a multiple of 5 is

n/m means that n is factor of m, then the relation 'f' is