What is the natural number n>=1 given A(n,1)+C(n,1)=10?

A is arrangements and C is combinations

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We'll write the formula that gives the arrangements of n elements taken 1 at a time.

A(n,1) = `(n!)/((n-1)!)`

Now, we'll write the formula that gives the number of combinations:

C(n,1) = `(n!)/(1!*(n-1)!)`

We'll re-write the equation:

A(n,1)+C(n,1)=10

`(n!)/((n-1)!) + (n!)/(1!*(n-1)!) = 10`

But 1! = 1

`(n!)/((n-1)!) + (n!)/((n-1)!) = 10`

`` `(2n!)/((n-1)!) = 10`

But n! = (n-1)!*n

`(2*(n-1)!*n)/((n-1)!) = 10`

We'll reduce like terms:

2n = 10 => n = `10/2` => n = 5

**Since the natural number n is larger than 1, then the solution of the given equation A(n,1)+C(n,1)=10 is n = 5.**

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