Question Number 55100 by aseerimad last updated on 17/Feb/19 Commented by aseerimad last updated on 17/Feb/19 how no.14 is done?.....pls help Commented by aseerimad last updated on 17/Feb/19 thank you sir. But can u please convert the answer to the form that I gave?…
Question Number 120431 by john santu last updated on 31/Oct/20 Commented by john santu last updated on 31/Oct/20 $${old}\:{unswered}\:{question} \\ $$ Commented by mr W…
Question Number 54860 by Tinkutara last updated on 13/Feb/19 Commented by Tinkutara last updated on 14/Feb/19 $${But}\:{answer}\:{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{n}−{k}^{\mathrm{2}} +{k}−\mathrm{2}\right) \\ $$ Answered by kaivan.ahmadi…
Question Number 54740 by gunawan last updated on 10/Feb/19 $${Prove}\:{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:{r}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\: \\ $$$$\mathrm{3}.\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+..+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} \\ $$ Answered by Kunal12588 last updated on 10/Feb/19…
Question Number 54698 by mr W last updated on 09/Feb/19 $${How}\:{many}\:{words}\:{with}\:{at}\:{least}\:\mathrm{2}\:{letters} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{from} \\ $$$${TINKUTARA}? \\ $$ Answered by afachri last updated on 09/Feb/19 Commented…
Question Number 120209 by benjo_mathlover last updated on 30/Oct/20 $${Peter}\:{has}\:\mathrm{12}\:{relatives}\:\left(\mathrm{5}\:{man}\:\&\:\mathrm{7}\:{woman}\right) \\ $$$${and}\:{his}\:{wife}\:{also}\:{has}\:\mathrm{12}\:{relatives} \\ $$$$\left(\mathrm{5}\:{woman}\:\&\mathrm{7}\:{man}\right).\:{They}\:{do}\:{not} \\ $$$${have}\:{common}\:{relatives}.\:{They}\:{decided} \\ $$$${to}\:{invite}\:\mathrm{12}\:{guests}\:,{six}\:{each}\:{of} \\ $$$${their}\:{relatives},\:{such}\:{that}\:{there} \\ $$$${are}\:{six}\:{man}\:{and}\:{six}\:{woman}\: \\ $$$${among}\:{the}\:{guests}.\:{How}\:{many} \\…
Question Number 120188 by benjo_mathlover last updated on 30/Oct/20 $${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:. \\ $$$${Prove}\:{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\begin{pmatrix}{\:\:{n}−{k}}\\{\lfloor\frac{{n}−{k}}{\mathrm{2}}\rfloor}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:{n}}\end{pmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54646 by gunawan last updated on 08/Feb/19 $$\mathrm{Such}\:\mathrm{That} \\ $$$$\mathrm{a}.\:_{{n}+\mathrm{1}} {C}_{{r}} =\frac{\left({n}+\mathrm{1}\right).\:_{{n}} {C}_{{r}} }{\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\mathrm{b}.\:_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{4}…} =_{{n}} {C}_{\mathrm{1}}…
Question Number 185659 by mr W last updated on 25/Jan/23 $${prove}\:{for}\:{r},\:{n}\:\in\:\mathbb{N} \\ $$$$\underset{{k}={r}} {\overset{{n}} {\sum}}\begin{pmatrix}{{k}}\\{{r}}\end{pmatrix}\:=\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix} \\ $$$$\left({Hockey}−{stick}\:{identity}\right) \\ $$ Commented by mr W last updated…
Question Number 185642 by Mingma last updated on 24/Jan/23 Commented by mr W last updated on 24/Jan/23 $$={C}_{\mathrm{4}+\mathrm{1}} ^{\mathrm{26}+\mathrm{1}} ={C}_{\mathrm{5}} ^{\mathrm{27}} =\mathrm{80730} \\ $$ Answered…