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show that p(m,1)+p(n,1)=p(m+n,1) for all...

show that `p(m,1)+p(n,1)=p(m+n,1)` for all positive integer m,n .

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To show that \( p(m, 1) + p(n, 1) = p(m+n, 1) \) for all positive integers \( m \) and \( n \), we will start by understanding the notation \( p(m, r) \), which represents the number of permutations of \( m \) objects taken \( r \) at a time. The formula for permutations is given by: \[ p(m, r) = \frac{m!}{(m-r)!} \] In our case, since \( r = 1 \), we can simplify this to: \[ p(m, 1) = \frac{m!}{(m-1)!} = m \] Similarly, we can write: \[ p(n, 1) = \frac{n!}{(n-1)!} = n \] Now we can substitute these values into the left-hand side of the equation: \[ p(m, 1) + p(n, 1) = m + n \] Next, we need to evaluate the right-hand side of the equation: \[ p(m+n, 1) = \frac{(m+n)!}{(m+n-1)!} = m+n \] Now we can compare both sides: \[ p(m, 1) + p(n, 1) = m + n \] \[ p(m+n, 1) = m + n \] Since both sides are equal, we can conclude that: \[ p(m, 1) + p(n, 1) = p(m+n, 1) \] Thus, we have shown that the equation holds true for all positive integers \( m \) and \( n \).
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