Let A a1a2a3 a4 and B b1 b2b3b4b5 so that A 4 and B 5 Ho

Let. A = {a_1,a_2,a_3, a_4} and B = {b_1, b_2,b_3,b_4,b_5} (so that \\A\\ = 4 and \\B\\ = -5). How many injective functions f : A rightarrow B satisfy f(a_1) = b_1 or f(a_2) = b_2? (This is an inclusive or.)

Solution

total number of required cases will be =(number of injective functions when f(a1)=b1)+ (number of injective functions when f(a2)=b2)- (number of injective functions when f(a1)=b1 and f(a2)=b2) ...(i)

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number of injective functions when f(a1)=b1

when f(a1)=b1 then a2 can be associated with any of b2,b3,b4 or b5 so that can be done in 4 ways

then a3 can be associated with any of remaining 3 letters

then a4 can be associated with any of remaining 2 letters

so number of ways =4x3x2 when a2 starts choosing

same way we can start with a3 or a4

so total number of injective functions when f(a1)=b1 is 3(4x3x2)

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number of injective functions when f(a2)=b2

when f(a2)=b2 then a1 can be associated with any of b1,b3,b4 or b5 so that can be done in 4 ways

then a3 can be associated with any of remaining 3 letters

then a4 can be associated with any of remaining 2 letters

so number of ways =4x3x2 when a2 starts choosing

same way we can start with a3 or a4

so total number of injective functions when f(a2)=b2 is 3(4x3x2)

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number of injective functions when f(a1)=b1 and f(a2)=b2

when f(a1)=b1 and f(a2)=b2 then a3 can be associated with any remaining 3 letters of b

then a4 can be associated with any of remaining 2 letters

so number of ways =3x2 when a3 starts choosing

same way we can start with a4

so total number of injective functions when f(a1)=b1 and f(a2)=b2 is 2(3x2)

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now plug these into formula (i)

we get total number of injective functions =3(4x3x2)+3(4x3x2)-2(3x2)

=126

Hence final answer is 126