If ${}^{n}C_{r-1}=28$, ${}^{n}C_{r}=56$, ${}^{n}C_{r+1}=70$, then the value of $n$ and $r$ are
(A) $n=8,\ r=3$
(B) $n=8,\ r=4$
(C) $n=9,\ r=4$
(D) $n=9,\ r=3$

Question

Accepted Answer

> [!TLDR]
> From $56/28=2$ and $70/56=5/4$ we get $n+1=3r$ and $4n=9r+5$, giving $r=3,\ n=8$.

### Step 1 — first ratio

$$\frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r}=\frac{56}{28}=2\ \Rightarrow\ n-r+1=2r\ \Rightarrow\ n+1=3r.$$

### Step 2 — second ratio

$$\frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1}=\frac{70}{56}=\frac54\ \Rightarrow\ 4(n-r)=5(r+1)\ \Rightarrow\ 4n=9r+5.$$

### Step 3 — solve

From Step 1, $n=3r-1$. Substitute:

$$4(3r-1)=9r+5\ \Rightarrow\ 12r-4=9r+5\ \Rightarrow\ 3r=9\ \Rightarrow\ r=3,\ n=8.$$

### Step 4 — verify

$${}^8C_2=28,\quad {}^8C_3=56,\quad {}^8C_4=70.\ \checkmark$$ …

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Step 1 — first ratio

Step 2 — second ratio

Step 3 — solve

Step 4 — verify

Step 1 — first ratio

Step 2 — second ratio

Step 3 — solve

Step 4 — verify

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