Question

how many arrangements of the digits 0123456789 are there that do not end with 8 and do not begin with3​

Answer 1

Answer:

let,

n(A)=arrangements ending with 8=9!

n(B)=arrangements ending with 3=9!

n(E)=total arrangements =10!

n(A⋂B)=8!

We require: n(E)−n(A⋃B) but n(A⋃B)=n(A)+n(B)−n(a⋂B)

Since A and B are not mutually exclusive:

n(E)−(n(A)+n(B)−n(A⋂B))

=10!−(9!+9!−8!)

Step-by-step explanation:

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