If A = 1,2,3,4 B = 1,2,3,4,5,6 and f: A arrow B is an injective mapping satisfying f(i) ≠ i, then number of such mappings are -

Question

If A = 1,2,3,4 B = 1,2,3,4,5,6 and f: A arrow B is an injective mapping satisfying f(i) ≠ i, then number of such mappings are - (A) 182 (B) 181 (C) 183 (D) None

Tardigrade Admin · Accepted Answer

Using inclusion and exclusion No. of functions =6 × 5 × 4 × 3- 4 C 1 ⋅ 5 × 4 × 3- 4 C 2.. .4 × 3+ 4 C 3 ⋅ 3- 4 C 4 =181

Q. If $A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6}$ and $f : A \to B$ is an injective mapping satisfying $f (i) \neq = i$ , then number of such mappings are -

182

181

183

None

Using inclusion and exclusion
No. of functions $= 6 \times 5 \times 4 \times 3 - {^{4} C_{1} \cdot 5 \times 4 \times 3 -^{4} C_{2}$ . $4 \times 3 +^{4} C_{3} \cdot 3 -^{4} C_{4}} = 181$

Q. If A={1,2,3,4},B={1,2,3,4,5,6} and f:A→B is an injective mapping satisfying f(i)=i, then number of such mappings are -

182

181

183

None

Using inclusion and exclusion No. of functions =6×5×4×3−{4C1​⋅5×4×3−4C2​. 4×3+4C3​⋅3−4C4​}=181

Q. If $A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6}$ and $f : A \to B$ is an injective mapping satisfying $f (i) \neq = i$ , then number of such mappings are -

Using inclusion and exclusion
No. of functions $= 6 \times 5 \times 4 \times 3 - {^{4} C_{1} \cdot 5 \times 4 \times 3 -^{4} C_{2}$ . $4 \times 3 +^{4} C_{3} \cdot 3 -^{4} C_{4}} = 181$