The total number of injections (one-one into | vedclass

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(D) Let $A = \{a_{1}, a_{2}, a_{3}, a_{4}\}$ and $B = \{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\}$.Here, the number of elements in set $A$ is

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(D) Let $A = \{a_{1}, a_{2}, a_{3}, a_{4}\}$ and $B = \{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\}$.Here, the number of elements in set $A$ is $n(A) = 4$ and the number of elements in set $B$ is $n(B) = 7$.An injection (one-one mapping) from set $A$ to set $B$ exists if we choose $4$ distinct elements from $B$ and assign them to the $4$ elements of $A$.The total number of such injections is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$, where $n = 7$ and $r = 4$.Total injections $= {}^{7}P_{4} = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 	imes 6 	imes 5 	imes 4 = 840$.

The total number of injections (one-one into mappings) from the set $A = \{a_{1}, a_{2}, a_{3}, a_{4}\}$ to the set $B = \{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\}$ is:

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