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

Question

Vedantu Content Team · Accepted Answer

Hint: We start the problem by expanding the given summation up to the terms given. We use the fact that ${}^{n}{{C}_{n}}=1$, for any positive value of n to replace ${}^{r+1}{{C}_{r+1}}$ in place of ${}^{r}{{C}_{r}}$. We use the fact ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$ 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 $\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}}$ and we need to find the value of x and y.we have $\sum\limits_{k=1}^{n-r}{{}^{n-k}{{C}_{r}}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{n-\left( n-r+1 \right)}{{C}_{r}}+{}^{n-\left( n-r \right)}{{C}_{r}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+1}{{C}_{r}}+{}^{r}{{C}_{r}}={}^{x}{{C}_{y}}$ ---(1).We know that ${}^{n}{{C}_{n}}=1$, for any positive value of n. so, we get $${}^{r}{{C}_{r}}={}^{r+1}{{C}_{r+1}}=1$$.We now substitute ${}^{r+1}{{C}_{r+1}}$ in place of ${}^{r}{{C}_{r}}$ in the equation (1).$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r}}+{}^{r+1}{{C}_{r}}+{}^{r+1}{{C}_{r+1}}={}^{x}{{C}_{y}}$---(2).We know that ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$. We use this result in equation (2).$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r}}+{}^{r+2}{{C}_{r+1}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+......+{}^{r+2}{{C}_{r+1}}={}^{x}{{C}_{y}}$.Similarly, this trend continues up to n–3 as shown.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-3}{{C}_{r}}+{}^{n-3}{{C}_{r+1}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-2}{{C}_{r}}+{}^{n-2}{{C}_{r+1}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n-1}{{C}_{r}}+{}^{n-1}{{C}_{r+1}}={}^{x}{{C}_{y}}$.$\Rightarrow {}^{n}{{C}_{r+1}}={}^{x}{{C}_{y}}$.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 ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ holds true. We can expect problems to get the value of the combination by giving values of n and r.

If ∑k=1n−rn−kCr=xCy, then find the values of x and y?(a) x=n+1; y=r,(b) x=n; y=r+1,(c) x=n; y=r,(d) x=n+1; y=r+1.

If $\sum_{k = 1}^{n - r}^{n - k} C_{r} =^{x} C_{y}$ , then find the values of x and y?
(a) $x = n + 1$ ; $y = r$ ,
(b) $x = n$ ; $y = r + 1$ ,
(c) $x = n$ ; $y = r$ ,
(d) $x = n + 1$ ; $y = r + 1$ .