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Question: Prove that \({}^n{c_r} + 2{}^n{c_{r - 1}} + {}^n{c_{r - 2}} = {}^{n + 2}{c_r}\)....

Prove that ncr+2ncr−1+ncr−2=n+2cr{}^n{c_r} + 2{}^n{c_{r - 1}} + {}^n{c_{r - 2}} = {}^{n + 2}{c_r}.

Explanation

Solution

We have to solve the question of permutation and combination by using the concepts and formulas. We have to show that the left hand side is equal to the right hand side. The terms represent different ways of selection of things.

Complete step by step solution:
According to the question we have to prove that ncr+2ncr−1+ncr−2=n+2cr{}^n{c_r} + 2{}^n{c_{r - 1}} + {}^n{c_{r - 2}} = {}^{n + 2}{c_r}
When is it written as ncr{}^n{c_r} it means the number of ways of combination of r different items out of a total of n different things.
So let us first solve the left hand side of the equation
We will use the formula of permutation and combination
⇒ncr+ncr+1=n+1cr+1\Rightarrow {}^n{c_r} + {}^n{c_{r + 1}} = {}^{n + 1}{c_{r + 1}}
Which represents that the addition of the total number of ways of combination of r different items out of a total of n different things with the total number of ways of combination of r+1 different items out of a total of n different things, and it will be equal to the total number of ways of combination of r+1 different items out of a total of n+1 different things.
So we can rewrite the equation as
⇒ncr+ncr−1+ncr−1+ncr−2\Rightarrow {}^n{c_r} + {}^n{c_{r - 1}} + {}^n{c_{r - 1}} + {}^n{c_{r - 2}}
⇒n+1cr+n+1cr−1\Rightarrow {}^{n + 1}{c_r} + {}^{n + 1}{c_{r - 1}}
Now, with the same formula it can be written as
⇒n+2cr\Rightarrow {}^{n + 2}{c_r}
Hence, it is equal to right hand side equation
Hence proved

Note:
We can observe that the formula represents different ways of selection of things from a collection. When is it written as ncr{}^n{c_r} it means the number of ways of combination of r different items out of a total of n different things. We should be concentrative towards these formulas as we can make many intakes there.