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Question Number 3273 by Yozzi last updated on 09/Dec/15

$$\ast \ast \ast \ast Onecanonlymovetothe$$$$\ast \ast \ast \ast rightordownwardsonthe$$$$\ast \ast \ast \ast 4by6pointlatticeshown.$$$$\ast \ast \ast \ast Howmanypathsfrom\ast to$$$$\ast \ast \ast \ast \ast arethere?$$$$\ast \ast \ast \ast$$$$$$

Answered by Filup last updated on 09/Dec/15

$$\mathrm{Assume}\mathrm{you}\mathrm{move}\ast$$$$$$$$\mathrm{You}\mathrm{must}\mathrm{move}:$$$$\mathrm{Down}5\mathrm{times}$$$$\mathrm{Right}3\mathrm{times}$$$$$$$$\mathrm{You}\mathrm{can}\mathrm{move}\mathrm{in}\mathrm{any}combination$$$$\mathrm{just}\mathrm{so}\mathrm{long}\mathrm{as}\mathrm{all}\mathrm{moves}\mathrm{are}\mathrm{made}$$$$$$$$\therefore \mathrm{it}\mathrm{is}\mathrm{the}\mathrm{permutation}{}^5{\mathrm{P}}_3=60$$

Commented by Filup last updated on 09/Dec/15

$$\mathrm{I}\mathrm{hope}\mathrm{this}\mathrm{is}\mathrm{correct}!$$

Answered by prakash jain last updated on 09/Dec/15

$$x_1{d}_1{x}_2{d}_2{x}_3{d}_3{x}_4{d}_4{x}_5{d}_5{x}_6$$$$xneedstobefilledwith{r}_1{r}_2{r}_3$$$$r_1{r}_2{r}_{3}together6choices$$$$r_1{r}_{2}and{r}_{3}5+4+3+2+1=15$$$$\left(r_1{r}_2\mathrm{at}{x}_1\mathrm{leaves}5\mathrm{possibilities}\mathrm{for}{r}_3\right)$$$$r_{1}and{r}_2{r}_{3}5+4+3+2+1=15$$$$\left(sameasabove\right)$$$$r_1{r}_2{r}_3$$$$r_{1}at{x}_1{x}_1{x}_2{x}_3{x}_1{x}_2{x}_4{x}_1{x}_2{x}_5{x}_1{x}_2{x}_6$$$$x_1{x}_3{x}_4{x}_1{x}_3{x}_5{x}_1{x}_3{x}_6$$$$x_1{x}_4{x}_5{x}_1{x}_5{x}_6$$$$x_1{x}_5{x}_6=4+3+2+1=10$$$$similarly{r}_{1}taking{x}_{2}willgive(3+2+1)=6$$$$similarly{r}_{1}taking{x}_{3}willgive(2+1)=3$$$$similarly{r}_{1}taking{x}_{4}willgive\left(1\right)=1$$$$total=6+15+15+10+6+3+1=56$$

Commented by prakash jain last updated on 09/Dec/15

$$\mathrm{Answer}\mathrm{in}\mathrm{factorials}$$$$\mathrm{csse}1:{}^6{P}_1\left(3\mathrm{r}\mathrm{together}\right)=6$$$$\mathrm{case}2:{}^6{P}_2\left(r2r2slotstobefilledin6\right)=30$$$$\mathrm{case}3:\frac{P_3}{3!}(rrr3slotstobefilled.3duplicates)=20$$

Commented by prakash jain last updated on 11/Dec/15

$$\mathrm{You}\mathrm{can}\mathrm{also}\mathrm{solve}\mathrm{this}\mathrm{problem}\mathrm{with}$$$$\mathrm{combination}\mathrm{formula}.$$$$8\mathrm{steps}\mathrm{need}\mathrm{to}\mathrm{be}\mathrm{taken}$$$$\mathrm{SSSSSSSS}(5\mathrm{D},3\mathrm{R})$$$$\mathrm{The}\mathrm{problem}\mathrm{is}\mathrm{choosing}3\mathrm{steps}\mathrm{among}$$$$8\mathrm{where}\mathrm{R}\mathrm{will}\mathrm{be}\mathrm{placed}\mathrm{remaining}$$$$\mathrm{position}\mathrm{will}\mathrm{have}\mathrm{D}.$$$$\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{ways}={}^8{C}_3=56$$

Answered by Rasheed Soomro last updated on 09/Dec/15

$$\mathcal{T}otalpositions24.\mathcal{L}eteveryposition$$$$bespecifiedbycoordinatesasbelow:$$$$00102030$$$$01112131$$$$02122232$$$$03132333$$$$04142434$$$$05152535$$$$\mathcal{F}rom003+5=8movesarepossible$$$$\mathcal{F}rommn$$

Commented by prakash jain last updated on 09/Dec/15

$$\mathrm{Consider}\mathrm{this}$$$$3\mathrm{r}+5\mathrm{d}$$$$1\mathrm{d}+3\mathrm{r}+4\mathrm{d}$$$$2\mathrm{d}+3\mathrm{r}+3\mathrm{d}$$$$3\mathrm{d}+3\mathrm{r}+2\mathrm{d}$$$$4\mathrm{d}+3\mathrm{r}+\mathrm{d}$$$$4\mathrm{d}+\mathrm{r}$$$$2\mathrm{r}+5\mathrm{d}+\mathrm{r}$$$$2\mathrm{r}+4\mathrm{d}+\mathrm{r}+\mathrm{d}$$$$2\mathrm{r}+3\mathrm{d}+\mathrm{r}+2\mathrm{d}$$$$2\mathrm{r}+2\mathrm{d}+\mathrm{r}+3\mathrm{d}$$$$2\mathrm{r}+\mathrm{d}+\mathrm{r}+4\mathrm{d}$$$$\mathrm{and}\mathrm{so}\mathrm{on}.$$

Commented by Rasheed Soomro last updated on 09/Dec/15

$$\mathcal{T}{h}^{\alpha }n{\mathcal{K}}^{\mathcal{S}}.$$

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