Undergrad Transforming an equation with logarithms

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To transform the logarithm of the binomial coefficient using the Stirling approximation, the initial expression n log n - m log m - (n - m) log(n - m) can be manipulated by adding and subtracting m log n. This adjustment allows for the reorganization of terms to achieve the desired form of (n - m) log(n/(n - m)) + m log(n/m). The key is to recognize how to group and factor the logarithmic terms appropriately. By carefully applying logarithmic properties, the transformation can be completed. This method effectively simplifies the approximation of the binomial coefficient's logarithm.
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I want to approximate the logarithm of the Binomial coefficient log (n!/ ((n - m)! m!) with the the Stirling approximation log x! ≈ x log x - x

I got

n log n - m log m - (n - m) log(n - m)

but I want

(n - m) log (n/(n - m)) + m log (n/m)

Can someone help how to transform the first equation into the latter?
 
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Try to add and subtract ##m \log{n}## in your first expresion
 
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Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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