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…
Question Number 120090 by bramlexs22 last updated on 29/Oct/20 $${Find}\:{the}\:{number}\:{of}\:{subsets}\:{of} \\ $$$$\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},…,\mathrm{2000}\:\right\}\:{the}\:{sum}\:{of}\: \\ $$$${whose}\:{elements}\:{is}\:{divisible}\:{by}\:\mathrm{5} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119896 by benjo_mathlover last updated on 27/Oct/20 Commented by bobhans last updated on 28/Oct/20 $$\left({ii}\right)\:\mathrm{5}\:{girls}\:{must}\:{be}\:{together}\:\Leftrightarrow\:\mathrm{5}!×\mathrm{8}! \\ $$ Answered by mr W last updated…
Question Number 54286 by Tinkutara last updated on 01/Feb/19 Commented by Tinkutara last updated on 01/Feb/19 Answers 4d 5b Commented by mr W last updated on 05/Feb/19…
Question Number 185350 by cortano1 last updated on 20/Jan/23 $$\:\:\:{C}_{\mathrm{4}} ^{\mathrm{4}} +{C}_{\mathrm{4}} ^{\mathrm{5}} +{C}_{\mathrm{4}} ^{\mathrm{6}} +…+{C}_{\mathrm{4}} ^{\mathrm{26}} \:=? \\ $$ Commented by mr W last…
Question Number 119747 by benjo_mathlover last updated on 26/Oct/20 $${Suppose}\:{that}\:\mathrm{7}\:{blue}\:{balls}\:,\:\mathrm{8}\:{red}\:{balls}\:{and}\:\mathrm{9}\:{green} \\ $$$${balls}\:{should}\:{be}\:{put}\:{into}\:{three}\:{boxes}\:{labeled} \\ $$$$\mathrm{1},\mathrm{2}\:{and}\:\mathrm{3},\:{so}\:{that}\:{any}\:{box}\:{contains}\:{at}\:{least} \\ $$$${one}\:{balls}\:{of}\:{each}\:{colour}.\:{How}\:{many}\:{ways} \\ $$$${can}\:{this}\:{arrangement}\:{be}\:{done}? \\ $$ Answered by mr W last…