تباديل وتوافيق

الملف الشخصي · 09-08-2014, 03:59 PM

السلام عليكم ورحمة الله وبركاته

الملف الشخصي · 09-08-2014, 05:06 PM

n=7 and r=3

العبارة لا تتحقق عند r=0 or 1 or 2

عند r=3

N(n-1)(n-2)=7*6*5
Then n=7

When r>3 العبارة لاتتحقق

الملف الشخصي · 09-09-2014, 03:01 PM

الملف الشخصي · 09-09-2014, 05:26 PM

السلام عليكم ورحمة الله وبركاته
وننتظر حلول اخرى بعد الحلين الرائعين لكل من الاستاذين المحترمين / صلاح احمد ومهيب الأرنؤوطى

الملف الشخصي · 09-28-2014, 08:32 PM

السلام عليكم ورحمة الله وبركاته
الحلان المطروحان لا يمثلان حلا رياضيا انما حلا تخمينيا

الملف الشخصي · 09-30-2014, 05:56 AM

${C_{r}^{n}}=\frac{P_{r}^{n}}{r!}\rightarrow {\color{Blue} P_{r}^{n}=r!\times C_{r}^{n}---\left ( 1 \right )}$

$P_{r}^{n}-C_{r}^{n}=175$

$r!\times C_{r}^{n}-C_{r}^{n}=175$

${\color{Blue} \left ( C_{r}^{n} \right )\left ( r!-1 \right )=5\times 5\times 7=5^{2}\times 7---\left ( 2 \right )}$

على أعتبار أن (n),(r) عددان طبيعيان من المعادلة (2) نستنتج أن:

$either\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175\right )} \right )or\left ( {\color{Blue} \left ( r!-1=5 \right )}and{\color{Blue} \left ( C_{r}^{n} =35\right )} \right )or\left ( {\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left (C_{r}^{n}=25 \right )} \right )$

$or\left ( {\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7\right )} \right )or\left ( {\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )} \right )or\left ( {\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left (C_{r}^{n}=1 \right )} \right )$

${\color{Red} \left ( 1 \right )}when\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175 \right )} \right )$

$r!-1=1\rightarrow r!=1+1\rightarrow r!=2\rightarrow r!=2!\rightarrow {\color{Blue} r=2}$

$C_{r}^{n}=175\rightarrow C_{2}^{n}=175\rightarrow \frac{\left n( n-1 \right )}{2!}=175\rightarrow n^{2}-n=350\rightarrow n^{2}-n-350=0$

$n=\frac{-b\mp \sqrt{b^{2}-4ac}}{2a}\rightarrow n=\frac{1\mp \sqrt{\left ( -1 \right )^{2}-4\times 1\times \left ( -350 \right )}}{2\times 1}$

$n=\frac{1\mp \sqrt{1+1400}}{2}\rightarrow n=\frac{1\mp \sqrt{1401}}{2}$

$either\left ( {\color{Blue} n=\frac{1+\sqrt{1401}}{2}}\notin N\left ( refuse \right )\right )or\left ( {\color{Blue} n=\frac{1-\sqrt{1401}}{2}}\notin N\left ( refuse \right ) \right )$

${\color{Red} \left ( 2 \right )}when\left ( {\color{Blue} \left ( r!-1=5 \right )} and{\color{Blue} \left ( C_{r}^{n}=35 \right )}\right )$

$r!-1=5\rightarrow r!=5+1\rightarrow r!=6\rightarrow r!=3!\rightarrow {\color{Blue} r=3}$

$C_{r}^{n}=35\rightarrow C_{3}^{n}=35\rightarrow \frac{n\left ( n-1 \right )\left ( n-2 \right )}{3!}=35$

$n\left ( n^{2}-3n+2 \right )=210\rightarrow n^{3}{\color{Red} -3n^{2}}+2n-210=0$

$\rightarrow \left ( n^{3}{\color{Red} -7n^{2}} \right )+\left ( {\color{Red} 4n^{2}}+2n-210 \right )=0$

$\rightarrow n^{2}\left ( n-7 \right )+\left ( n-7 \right )\left ( 4n+30 \right )=0$

$\rightarrow \left ( n-7 \right )\left ( n^{2}+4n+30 \right )=0\rightarrow either\left ( n-7 \right )=0\rightarrow {\color{Blue} n=7}$

$or\left ( n^{2}+4n{\color{Red} +}{\color{Red} 30}\right )=0\rightarrow \left ( n^{2}+4n{\color{Red} +4 }\right ){\color{Red} +26}=0$

$\rightarrow \left ( n+2 \right )^{2}+26=0\left ( refuse \right )\left ( because{\color{Blue} \left ( n+2 \right )^{2}+26> 0\ }\forall {\color{Blue} \ n\in N} \right )$

${\color{Red} \left ( 3 \right )}when{\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left ( C_{r}^{n}=25 \right )}$

$r!-1=7\rightarrow r!=7+1\rightarrow r!=8\left ( but{\color{Blue} \left ( 6< 8< 24 \right )}\rightarrow {\color{Blue} \left ( 3!< 8<4! \right )} \right )$

$\therefore {\color{Blue} r!=8}\left ( refuse \right )$

${\color{Red} \left ( 4 \right )}when{\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7 \right )}$

$r!-1=25\rightarrow r!=25+1\rightarrow {\color{Blue} r!=26}\left ( but{\color{Blue} \left ( 24< 26< 120 \right )}\rightarrow {\color{Blue} \left ( 4!< 26< 5! \right )} \right )$

$\therefore {\color{Blue} r!=26}\left ( refuse \right )$

${\color{Red} \left ( 5 \right )}when{\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )}$

$r!-1=35\rightarrow r!=35+1\rightarrow {\color{Blue} r!=36}\left ( but{\color{Blue} \left ( 24< 36< 120 \right )\rightarrow {\color{Blue} \left ( 4!< 36< 5! \right )}} \right )$

$\therefore {\color{Blue} r!=36}\left ( refuse \right )$

${\color{Red} \left ( 6 \right )}when{\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left ( C_{r}^{n} =1\right )}$

$r!-1=175\rightarrow r!=175+1\rightarrow {\color{Blue} r!=176}\left ( but{\color{Blue} \left ( 120> 176> 720 \right )}\rightarrow {\color{Blue} \left ( 5!> 176> 6! \right )} \right )$

$\therefore {\color{Blue} r!=176}\left ( refuse \right )$

الحالة الصحيحة الوحيدة هي:

${\color{Blue} r=3},{\color{Blue} n=7}$

الملف الشخصي · 09-30-2014, 06:05 AM

${C_{r}^{n}}=\frac{P_{r}^{n}}{r!}\rightarrow {\color{Blue} P_{r}^{n}=r!\times C_{r}^{n}---\left ( 1 \right )}$

$P_{r}^{n}-C_{r}^{n}=175$

$r!\times C_{r}^{n}-C_{r}^{n}=175$

${\color{Blue} \left ( C_{r}^{n} \right )\left ( r!-1 \right )=5\times 5\times 7=5^{2}\times 7---\left ( 2 \right )}$

على أعتبار أن (n),(r) عددان طبيعيان من المعادلة (2) نستنتج أن:

$either\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175\right )} \right )or\left ( {\color{Blue} \left ( r!-1=5 \right )}and{\color{Blue} \left ( C_{r}^{n} =35\right )} \right )or\left ( {\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left (C_{r}^{n}=25 \right )} \right )$

$or\left ( {\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7\right )} \right )or\left ( {\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )} \right )or\left ( {\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left (C_{r}^{n}=1 \right )} \right )$

${\color{Red} \left ( 1 \right )}when\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175 \right )} \right )$

$r!-1=1\rightarrow r!=1+1\rightarrow r!=2\rightarrow r!=2!\rightarrow {\color{Blue} r=2}$

$C_{r}^{n}=175\rightarrow C_{2}^{n}=175\rightarrow \frac{\left n( n-1 \right )}{2!}=175\rightarrow n^{2}-n=350\rightarrow n^{2}-n-350=0$

$n=\frac{-b\mp \sqrt{b^{2}-4ac}}{2a}\rightarrow n=\frac{1\mp \sqrt{\left ( -1 \right )^{2}-4\times 1\times \left ( -350 \right )}}{2\times 1}$

$n=\frac{1\mp \sqrt{1+1400}}{2}\rightarrow n=\frac{1\mp \sqrt{1401}}{2}$

$either\left ( {\color{Blue} n=\frac{1+\sqrt{1401}}{2}}\notin N\left ( refuse \right )\right )or\left ( {\color{Blue} n=\frac{1-\sqrt{1401}}{2}}\notin N\left ( refuse \right ) \right )$

${\color{Red} \left ( 2 \right )}when\left ( {\color{Blue} \left ( r!-1=5 \right )} and{\color{Blue} \left ( C_{r}^{n}=35 \right )}\right )$

$r!-1=5\rightarrow r!=5+1\rightarrow r!=6\rightarrow r!=3!\rightarrow {\color{Blue} r=3}$

$C_{r}^{n}=35\rightarrow C_{3}^{n}=35\rightarrow \frac{n\left ( n-1 \right )\left ( n-2 \right )}{3!}=35$

$n\left ( n^{2}-3n+2 \right )=210\rightarrow n^{3}{\color{Red} -3n^{2}}+2n-210=0$

$\rightarrow \left ( n^{3}{\color{Red} -7n^{2}} \right )+\left ( {\color{Red} 4n^{2}}+2n-210 \right )=0$

$\rightarrow n^{2}\left ( n-7 \right )+\left ( n-7 \right )\left ( 4n+30 \right )=0$

$\rightarrow \left ( n-7 \right )\left ( n^{2}+4n+30 \right )=0\rightarrow either\left ( n-7 \right )=0\rightarrow {\color{Blue} n=7}$

$or\left ( n^{2}+4n{\color{Red} +}{\color{Red} 30}\right )=0\rightarrow \left ( n^{2}+4n{\color{Red} +4 }\right ){\color{Red} +26}=0$

$\rightarrow \left ( n+2 \right )^{2}+26=0\left ( refuse \right )\left ( because{\color{Blue} \left ( n+2 \right )^{2}+26> 0\ }\forall {\color{Blue} \ n\in N} \right )$

${\color{Red} \left ( 3 \right )}when{\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left ( C_{r}^{n}=25 \right )}$

$r!-1=7\rightarrow r!=7+1\rightarrow r!=8\left ( but{\color{Blue} \left ( 6< 8< 24 \right )}\rightarrow {\color{Blue} \left ( 3!< 8<4! \right )} \right )$

$\therefore {\color{Blue} r!=8}\left ( refuse \right )$

${\color{Red} \left ( 4 \right )}when{\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7 \right )}$

$r!-1=25\rightarrow r!=25+1\rightarrow {\color{Blue} r!=26}\left ( but{\color{Blue} \left ( 24< 26< 120 \right )}\rightarrow {\color{Blue} \left ( 4!< 26< 5! \right )} \right )$

$\therefore {\color{Blue} r!=26}\left ( refuse \right )$

${\color{Red} \left ( 5 \right )}when{\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )}$

$r!-1=35\rightarrow r!=35+1\rightarrow {\color{Blue} r!=36}\left ( but{\color{Blue} \left ( 24< 36< 120 \right )\rightarrow {\color{Blue} \left ( 4!< 36< 5! \right )}} \right )$

$\therefore {\color{Blue} r!=36}\left ( refuse \right )$

${\color{Red} \left ( 6 \right )}when{\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left ( C_{r}^{n} =1\right )}$

$r!-1=175\rightarrow r!=175+1\rightarrow {\color{Blue} r!=176}\left ( but{\color{Blue} \left ( 120< 176< 720 \right )}\rightarrow {\color{Blue} \left ( 5!< 176< 6! \right )} \right )$

$\therefore {\color{Blue} r!=176}\left ( refuse \right )$

الحالة الصحيحة الوحيدة هي:

${\color{Blue} r=3},{\color{Blue} n=7}$

الملف الشخصي · 09-30-2014, 06:10 AM

أرجو من الأساتذة الكرام حذف المشاركة الأولى لحصول خطأ فيها ولعدم معرفتي بحذفها أو تعديلها مع أبقاء المشاركة الثانية هذا ولكم جزيل الشكر والتقدير

الملف الشخصي · 09-30-2014, 06:18 AM

${C_{r}^{n}}=\frac{P_{r}^{n}}{r!}\rightarrow {\color{Blue} P_{r}^{n}=r!\times C_{r}^{n}---\left ( 1 \right )}$

$P_{r}^{n}-C_{r}^{n}=175$

$r!\times C_{r}^{n}-C_{r}^{n}=175$

${\color{Blue} \left ( C_{r}^{n} \right )\left ( r!-1 \right )=5\times 5\times 7=5^{2}\times 7---\left ( 2 \right )}$

على أعتبار أن (n),(r) عددان طبيعيان من المعادلة (2) نستنتج أن:

$either\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175\right )} \right )or\left ( {\color{Blue} \left ( r!-1=5 \right )}and{\color{Blue} \left ( C_{r}^{n} =35\right )} \right )or\left ( {\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left (C_{r}^{n}=25 \right )} \right )$

$or\left ( {\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7\right )} \right )or\left ( {\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )} \right )or\left ( {\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left (C_{r}^{n}=1 \right )} \right )$

${\color{Red} \left ( 1 \right )}when\left ( {\color{Blue} \left ( r!-1=1 \right )}and{\color{Blue} \left ( C_{r}^{n}=175 \right )} \right )$

$r!-1=1\rightarrow r!=1+1\rightarrow r!=2\rightarrow r!=2!\rightarrow {\color{Blue} r=2}$

$C_{r}^{n}=175\rightarrow C_{2}^{n}=175\rightarrow \frac{\left n( n-1 \right )}{2!}=175\rightarrow n^{2}-n=350\rightarrow n^{2}-n-350=0$

$n=\frac{-b\mp \sqrt{b^{2}-4ac}}{2a}\rightarrow n=\frac{1\mp \sqrt{\left ( -1 \right )^{2}-4\times 1\times \left ( -350 \right )}}{2\times 1}$

$n=\frac{1\mp \sqrt{1+1400}}{2}\rightarrow n=\frac{1\mp \sqrt{1401}}{2}$

$either\left ( {\color{Blue} n=\frac{1+\sqrt{1401}}{2}}\notin N\left ( refuse \right )\right )or\left ( {\color{Blue} n=\frac{1-\sqrt{1401}}{2}}\notin N\left ( refuse \right ) \right )$

${\color{Red} \left ( 2 \right )}when\left ( {\color{Blue} \left ( r!-1=5 \right )} and{\color{Blue} \left ( C_{r}^{n}=35 \right )}\right )$

$r!-1=5\rightarrow r!=5+1\rightarrow r!=6\rightarrow r!=3!\rightarrow {\color{Blue} r=3}$

$C_{r}^{n}=35\rightarrow C_{3}^{n}=35\rightarrow \frac{n\left ( n-1 \right )\left ( n-2 \right )}{3!}=35$

$n\left ( n^{2}-3n+2 \right )=210\rightarrow n^{3}{\color{Red} -3n^{2}}+2n-210=0$

$\rightarrow \left ( n^{3}{\color{Red} -7n^{2}} \right )+\left ( {\color{Red} 4n^{2}}+2n-210 \right )=0$

$\rightarrow n^{2}\left ( n-7 \right )+\left ( n-7 \right )\left ( 4n+30 \right )=0$

$\rightarrow \left ( n-7 \right )\left ( n^{2}+4n+30 \right )=0\rightarrow either\left ( n-7 \right )=0\rightarrow {\color{Blue} n=7}$

$or\left ( n^{2}+4n{\color{Red} +}{\color{Red} 30}\right )=0\rightarrow \left ( n^{2}+4n{\color{Red} +4 }\right ){\color{Red} +26}=0$

$\rightarrow \left ( n+2 \right )^{2}+26=0\left ( refuse \right )\left ( because{\color{Blue} \left ( n+2 \right )^{2}+26> 0\ }\forall {\color{Blue} \ n\in N} \right )$

${\color{Red} \left ( 3 \right )}when{\color{Blue} \left ( r!-1=7 \right )}and{\color{Blue} \left ( C_{r}^{n}=25 \right )}$

$r!-1=7\rightarrow r!=7+1\rightarrow r!=8\left ( but{\color{Blue} \left ( 6< 8< 24 \right )}\rightarrow {\color{Blue} \left ( 3!< 8<4! \right )} \right )$

$\therefore {\color{Blue} r!=8}\left ( refuse \right )$

${\color{Red} \left ( 4 \right )}when{\color{Blue} \left ( r!-1=25 \right )}and{\color{Blue} \left ( C_{r}^{n}=7 \right )}$

$r!-1=25\rightarrow r!=25+1\rightarrow {\color{Blue} r!=26}\left ( but{\color{Blue} \left ( 24< 26< 120 \right )}\rightarrow {\color{Blue} \left ( 4!< 26< 5! \right )} \right )$

$\therefore {\color{Blue} r!=26}\left ( refuse \right )$

${\color{Red} \left ( 5 \right )}when{\color{Blue} \left ( r!-1=35 \right )}and{\color{Blue} \left ( C_{r}^{n} =5\right )}$

$r!-1=35\rightarrow r!=35+1\rightarrow {\color{Blue} r!=36}\left ( but{\color{Blue} \left ( 24< 36< 120 \right )\rightarrow {\color{Blue} \left ( 4!< 36< 5! \right )}} \right )$

$\therefore {\color{Blue} r!=36}\left ( refuse \right )$

${\color{Red} \left ( 6 \right )}when{\color{Blue} \left ( r!-1=175 \right )}and{\color{Blue} \left ( C_{r}^{n} =1\right )}$

$r!-1=175\rightarrow r!=175+1\rightarrow {\color{Blue} r!=176}\left ( but{\color{Blue} \left ( 120< 176< 720 \right )}\rightarrow {\color{Blue} \left ( 5!< 176< 6! \right )} \right )$

$\therefore {\color{Blue} r!=176}\left ( refuse \right )$

الحالة الصحيحة الوحيدة هي:

${\color{Blue} r=3},{\color{Blue} n=7}$

الملف الشخصي · 09-30-2014, 06:34 AM

أرجو من الأساتذة الكرام بحذف المشاركة الأولى والثانية وأبقاء المشاركة الأخيرة فقد قمت بالتعديل على المشاركة الخاطئة فعدلت ولكن أضيفت كمشاركة أخرى مع بقاء المشاركة الخاطئة هذا ولكم جزيل الشكر والتقدير

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