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Question: If \(\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}}\), then find the values of x and y? ...

If ∑k=1n−rn−kCr=xCy\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}}, then find the values of x and y?
(a) x=n+1x=n+1; y=ry=r,
(b) x=nx=n; y=r+1y=r+1,
(c) x=nx=n; y=ry=r,
(d) x=n+1x=n+1; y=r+1y=r+1.

Explanation

Solution

We start the problem by expanding the given summation up to the terms given. We use the fact
that nCn=1{}^{n}{{C}_{n}}=1, for any positive value of n to replace r+1Cr+1{}^{r+1}{{C}_{r+1}} in place of rCr{}^{r}{{C}_{r}}. We use the fact nCr+nCr−1=n+1Cr{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} to proceed through the problem. We do this step continuously to get the total sum into a single combination. We then compare combinations of both sides to get the values of x and y.

Complete step by step answer:
According to the problem, we have ∑k=1n−rn−kCr=xCy\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}} and we need to find the value of x and y.
we have ∑k=1n−rn−kCr=xCy\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}}.
⇒n−1Cr+n−2Cr+n−3Cr+......+n−(n−r+1)Cr+n−(n−r)Cr=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{n-\left( n-r+1 \right)}{{C}_{r}}+{}^{n-\left( n-r \right)}{{C}_{r}}={}^{x}{{C}_{y}}.
⇒n−1Cr+n−2Cr+n−3Cr+......+r+1Cr+rCr=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+1}{{C}_{r}}+{}^{r}{{C}_{r}}={}^{x}{{C}_{y}} ---(1).
We know that nCn=1{}^{n}{{C}_{n}}=1, for any positive value of n. so, we get rCr=r+1Cr+1=1{}^{r}{{C}_{r}}={}^{r+1}{{C}_{r+1}}=1.
We now substitute r+1Cr+1{}^{r+1}{{C}_{r+1}} in place of rCr{}^{r}{{C}_{r}} in the equation (1).
⇒n−1Cr+n−2Cr+n−3Cr+......+r+2Cr+r+1Cr+r+1Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r}}+{}^{r+1}{{C}_{r}}+{}^{r+1}{{C}_{r+1}}={}^{x}{{C}_{y}}---(2).
We know that nCr+nCr−1=n+1Cr{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}. We use this result in equation (2).
⇒n−1Cr+n−2Cr+n−3Cr+......+r+2Cr+r+2Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r}}+{}^{r+2}{{C}_{r+1}}={}^{x}{{C}_{y}}.
⇒n−1Cr+n−2Cr+n−3Cr+......+r+2Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r+1}}={}^{x}{{C}_{y}}.
Similarly, this trend continues up to n–3 as shown.
⇒n−1Cr+n−2Cr+n−3Cr+n−3Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+{}^{n-3}{{C}_{r+1}}={}^{x}{{C}_{y}}.
⇒n−1Cr+n−2Cr+n−2Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-2}{{C}_{r+1}}={}^{x}{{C}_{y}}.
⇒n−1Cr+n−1Cr+1=xCy\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-1}{{C}_{r+1}}={}^{x}{{C}_{y}}.
⇒nCr+1=xCy\Rightarrow {}^{n}{{C}_{r+1}}={}^{x}{{C}_{y}}.
We compare the places x and y on both sides and we get the values of x and y as n and r+1.
We have found the values of x and y as n and r+1.

So, the correct answer is “Option B”.

Note: We can also solve this by expanding the combination and taking the common elements between the combinations. We can also expect multiple answers for this type of problem as we know the property of combination nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}} holds true. We can expect problems to get the value of the combination by giving values of n and r.