<s>[INST] Pascals Identity Let there be a group of n boxers and we want to select k people out of it, suppose one of the persons name is ‘Prem’ , so no of ways to choose k people = (combinations in which Prem is present + combinations in which Prem is not present) i.e $ {n-1} \choose {k-1} $ + $ {n-1} \choose k $, this is equal to $ n \choose k $, this was given in book. $ \qquad \qquad \quad $ However what I did was that to assume we know two people among them named ‘Prem’ and ‘Ram’ so now no of combinations = (combinations in which both are present + combination in which Ram is present but not Prem + combination in which Prem is present but not Ram + combination in which none of them are present ) =  $ {n-2} \choose {k-2} $ + 2 $ {n-1} \choose {k-1} $ + $ {n-2} \choose {k} $  , this also should be equal to $ n \choose k $ , however when i tried with a value namely $ 6 \choose 4 $ , i didn’t get the two values same, what did i do wrong?
 [/INST] Short answer: Your middle term should be $2\binom{n-2}{k-1}$.
Explanation: If, from $n$ people—Prem, Ram, and $n-2$ others—you must choose Prem but not Ram, and $k-1$ others, then the $k-1$ are drawn from the last group of $n-2$ others.  Likewise if you choose Rem but not Pram, and $k-1$ others.  Both these terms are $\binom{n-2}{k-1}$; hence, the middle term should be $2\binom{n-2}{k-1}$.
For $n = 6, k = 4$, we have
$$
\binom{4}{2} + 2\binom{4}{3} + \binom{4}{4} = 6 + 2(4) + 1 = 15 = \binom{6}{4}
$$
Note that your identity can also be derived from the first identity, applied recursively:
$$
\begin{align}
\binom{n}{k} & = \binom{n-1}{k-1} + \binom{n-1}{k} \\
             & = \binom{n-2}{k-2} + \binom{n-2}{k-1}
               + \binom{n-2}{k-1} + \binom{n-2}{k} \\
             & = \binom{n-2}{k-2} + 2\binom{n-2}{k-1} + \binom{n-2}{k}
\end{align}
$$
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