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Exercise 6.4 · Q1

Q.If 18Cr=18Cr+2^{18}C_r = {}^{18}C_{r+2}, find rC5^rC_5

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The identity nCa=nCb^{n}C_{a}={}^{n}C_{b} forces a+b=na+b=n (when a≠ba\ne b); apply it to find rr, then compute rC5^{r}C_{5}.

[!FORMULA] nCr=n!r!(n−r)!^{n}C_{r}=\dfrac{n!}{r!(n-r)!}, and nCr=nCn−r^{n}C_{r}={}^{n}C_{n-r}. Consequently nCa=nCb^{n}C_{a}={}^{n}C_{b} implies either a=ba=b or a+b=na+b=n.

  1. Given 18Cr=18Cr+2^{18}C_{r}={}^{18}C_{r+2} with n=18n=18.
  2. Since r≠r+2r\ne r+2 for any real rr, the equality must come from r+(r+2)=18r+(r+2)=18.
  3. 2r+2=18⇒2r=16⇒r=82r+2=18\Rightarrow 2r=16\Rightarrow r=8.
  4. Now evaluate rC5=8C5=8!5! (8−5)!=8!5! 3!^{r}C_{5}={}^{8}C_{5}=\dfrac{8!}{5!\,(8-5)!}=\dfrac{8!}{5!\,3!}.
  5. 8!5! 3!=8×7×63×2×1=3366=56\dfrac{8!}{5!\,3!}=\dfrac{8\times7\times6}{3\times2\times1}=\dfrac{336}{6}=56.
  6. Self-check: r=8≤18r=8\le18, valid; and 18C8=18C10^{18}C_{8}={}^{18}C_{10} (since 8+10=188+10=18), confirming r+2=10r+2=10 is consistent with the property used.
✓Final answer

rC5=8C5=56^{r}C_{5}={}^{8}C_{5}=56

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