Q49 / 160 · WBJEE 2012 · Mathematics
The total number of injections (one-one into mappings) from $\quad\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right)$ is
- A400
- B420
- C800
- Correct answerD840
Solution
Let $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\begin{aligned} B &=\left\{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right\} \\ n(A) &=4, n(B)=7 \end{aligned}$
The total number of infections $={ }^{7} P_{4}$ $=\frac{71}{3 !}=7 \cdot 6 \cdot 5 \cdot 4=840$
Hence, the correct answer is option: 'D'.