If the number of bijective functions from a set A to set B is 120,the n(A)+n(B) is equal to
Bijective function is both a one-to-one or injective function, and an onto or surjective function. This means that for every function f: A \(\rightarrow\) B, each member a of domain A maps to precisely one unique member b of codomain B.

There are certain features that make a bijective function:
For fn to be bijective , n(A) = n(B) = n and no. of fn are = n! n! = 120
∴ n = 5 ∴ n(A) + n(B) = 10
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
Let \(M = \{1, 2, 3, ....., 16\}\), if a relation R defined on set M such that R = \((x, y) : 4y = 5x – 3, x, y (\in) M\). How many elements should be added to R to make it symmetric.