Set A consists of 6 different elements and set B consists of 4 different elements. Number of mappings which can be defined from the set A arrow B which are surjective, is

Question

Set A consists of 6 different elements and set B consists of 4 different elements. Number of mappings which can be defined from the set A arrow B which are surjective, is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Groups 1,1,1,3 or 1,1,2,2 Number of mappings =(6 ! 4 !/3 ! 3 !)+(6 ! 4 !/2 ! 2 ! 2 ! 2 !) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a45e9b8722031dcf933a2cc69344843a-.png" /> =480+1080=1560

Q. Set $A$ consists of $6$ different elements and set $B$ consists of $4$ different elements. Number of mappings which can be defined from the set $A \to B$ which are surjective, is

Groups $1, 1, 1, 3$
or $1, 1, 2, 2$
Number of mappings
$= \frac{6 ! 4 !}{3 ! 3 !} + \frac{6 ! 4 !}{2 ! 2 ! 2 ! 2 !}$

$= 480 + 1080 = 1560$

Q. Set A consists of 6 different elements and set B consists of 4 different elements. Number of mappings which can be defined from the set A→B which are surjective, is

Groups 1,1,1,3 or 1,1,2,2 Number of mappings =3!3!6!4!​+2!2!2!2!6!4!​ =480+1080=1560

Q. Set $A$ consists of $6$ different elements and set $B$ consists of $4$ different elements. Number of mappings which can be defined from the set $A \to B$ which are surjective, is

Groups $1, 1, 1, 3$
or $1, 1, 2, 2$
Number of mappings
$= \frac{6 ! 4 !}{3 ! 3 !} + \frac{6 ! 4 !}{2 ! 2 ! 2 ! 2 !}$

$= 480 + 1080 = 1560$