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Category: Permutation and Combination

shown-that-n-k-0-n-1-k-k-n-n-k-n-please-help-me-

Question Number 120954 by cantor last updated on 04/Nov/20 $$\boldsymbol{{shown}}\:\boldsymbol{{that}} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{n}}!=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\left(−\mathrm{1}\right)^{\boldsymbol{{k}}} \left(_{\:\boldsymbol{{k}}} ^{\:\boldsymbol{{n}}} \right)\left(\boldsymbol{{n}}−\boldsymbol{{k}}\right)^{\boldsymbol{{n}}} \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}}\:\boldsymbol{{me}} \\ $$ Terms of…

How-many-numbers-divisible-by-5-can-be-made-with-the-digits-2-3-4-and-5-where-no-digit-is-being-used-more-than-once-in-each-number-

Question Number 55119 by pieroo last updated on 17/Feb/19 $$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers},\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5},\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{made}\:\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{and}\:\mathrm{5}\:\mathrm{where} \\ $$$$\mathrm{no}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{being}\:\mathrm{used}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{number}? \\ $$ Commented by Tawa1 last updated on 17/Feb/19…

Show-that-for-n-N-r-0-n-P-r-n-n-e-where-x-denotes-the-greatest-integer-x-and-P-r-n-n-n-r-

Question Number 55099 by Joel578 last updated on 17/Feb/19 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:{n}\:\in\:\mathbb{N}, \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:{P}_{{r}} ^{{n}} \:=\:\lfloor{n}!\:{e}\rfloor \\ $$$$\mathrm{where}\:\lfloor{x}\rfloor\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\leqslant\:{x} \\ $$$$\mathrm{and}\:{P}_{{r}} ^{{n}} \:=\:\frac{{n}!}{\left({n}\:−\:{r}\right)!} \\ $$ Answered…

Prove-that-1-n-r-n-1-r-n-1-r-1-2-n-r-n-r-1-n-1-r-3-n-0-n-1-n-2-n-n-2-n-

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…