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One-can-only-move-to-the-right-or-downwards-on-the-4-by-6-point-lattice-shown-How-many-paths-from-to-are-there-

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Question Number 3273 by Yozzi last updated on 09/Dec/15
∗ ∗ ∗ ∗ One can only move to the ∗ ∗ ∗ ∗ right or downwards on the ∗ ∗ ∗ ∗ 4 by 6 point lattice shown. ∗ ∗ ∗ ∗ How many paths from ∗ to ∗ ∗ ∗ ∗ ∗ are there? ∗ ∗ ∗ ∗
$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\:\:{One}\:{can}\:{only}\:{move}\:{to}\:{the} \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\:\:{right}\:{or}\:{downwards}\:{on}\:{the} \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\:\:\mathrm{4}\:{by}\:\mathrm{6}\:{point}\:{lattice}\:{shown}. \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\:\:{How}\:{many}\:{paths}\:{from}\:\ast\:{to} \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast\:\:\:\:\:\:\ast\:{are}\:{there}?\: \\ $$$$\ast\:\:\ast\:\:\ast\:\:\ast \\ $$$$ \\ $$
Answered by Filup last updated on 09/Dec/15
Assume you move ∗ You must move: Down 5 times Right 3 times You can move in any combination just so long as all moves are made ∴it is the permutation^5 P_3 =60
$$\mathrm{Assume}\:\mathrm{you}\:\mathrm{move}\:\ast \\ $$$$ \\ $$$$\mathrm{You}\:\mathrm{must}\:\mathrm{move}: \\ $$$$\mathrm{Down}\:\mathrm{5}\:\mathrm{times} \\ $$$$\mathrm{Right}\:\mathrm{3}\:\mathrm{times} \\ $$$$ \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{move}\:\mathrm{in}\:\mathrm{any}\:{combination} \\ $$$$\mathrm{just}\:\mathrm{so}\:\mathrm{long}\:\mathrm{as}\:\mathrm{all}\:\mathrm{moves}\:\mathrm{are}\:\mathrm{made} \\ $$$$ \\ $$$$\therefore\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{permutation}\:^{\mathrm{5}} \mathrm{P}_{\mathrm{3}} =\mathrm{60} \\ $$
Commented by Filup last updated on 09/Dec/15
I hope this is correct!
$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{is}\:\mathrm{correct}! \\ $$
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 x needs to be filled with r_1 r_2 r_3 r_1 r_2 r_3 together 6 choices r_1 r_2 and r_3 5+4+3+2+1=15 (r_1 r_2 at x_1 leaves 5 possibilities for r_3 ) r_1 and r_2 r_3 5+4+3+2+1=15 (same as above) 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 will give (3+2+1)=6 similarly r_1 taking x_3 will give (2+1)=3 similarly r_1 taking x_4 will give (1)=1 total=6+15+15+10+6+3+1=56
$${x}_{\mathrm{1}} {d}_{\mathrm{1}} {x}_{\mathrm{2}} {d}_{\mathrm{2}} {x}_{\mathrm{3}} {d}_{\mathrm{3}} {x}_{\mathrm{4}} {d}_{\mathrm{4}} {x}_{\mathrm{5}} {d}_{\mathrm{5}} {x}_{\mathrm{6}} \\ $$$${x}\:{needs}\:{to}\:{be}\:{filled}\:{with}\:{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} \:{together}\:\:\:\:\mathrm{6}\:{choices} \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} \:{and}\:{r}_{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{15} \\ $$$$\:\:\:\:\:\left({r}_{\mathrm{1}} {r}_{\mathrm{2}} \:\mathrm{at}\:{x}_{\mathrm{1}} \:\mathrm{leaves}\:\mathrm{5}\:\mathrm{possibilities}\:\mathrm{for}\:{r}_{\mathrm{3}} \right) \\ $$$${r}_{\mathrm{1}} \:{and}\:{r}_{\mathrm{2}} {r}_{\mathrm{3}} \:\:\:\:\:\:\:\:\mathrm{5}+\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{15} \\ $$$$\:\:\:\:\:\:\left({same}\:{as}\:{above}\right) \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} \\ $$$${r}_{\mathrm{1}} \:{at}\:{x}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} \:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{4}} \:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{5}} \:\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} \:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{3}} {x}_{\mathrm{5}} \:\:{x}_{\mathrm{1}} {x}_{\mathrm{3}} {x}_{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} \:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{5}} {x}_{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} {x}_{\mathrm{5}} {x}_{\mathrm{6}} =\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}=\mathrm{10} \\ $$$${similarly}\:{r}_{\mathrm{1}} \:{taking}\:{x}_{\mathrm{2}} \:{will}\:{give}\:\left(\mathrm{3}+\mathrm{2}+\mathrm{1}\right)=\mathrm{6} \\ $$$${similarly}\:{r}_{\mathrm{1}} \:{taking}\:{x}_{\mathrm{3}} \:{will}\:{give}\:\left(\mathrm{2}+\mathrm{1}\right)=\mathrm{3} \\ $$$${similarly}\:{r}_{\mathrm{1}} \:{taking}\:{x}_{\mathrm{4}} \:{will}\:{give}\:\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${total}=\mathrm{6}+\mathrm{15}+\mathrm{15}+\mathrm{10}+\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{56} \\ $$
Commented by prakash jain last updated on 09/Dec/15
Answer in factorials csse 1:^6 P_1 (3r together)=6 case 2:^6 P_2 (r 2r 2 slots to be filled in 6)=30 case 3: ((P_3 )/(3!)) (r r r 3 slots to be filled. 3 duplicates)=20
$$\mathrm{Answer}\:\mathrm{in}\:\mathrm{factorials} \\ $$$$\:\mathrm{csse}\:\mathrm{1}:\:^{\mathrm{6}} {P}_{\mathrm{1}} \:\:\:\left(\mathrm{3r}\:\mathrm{together}\right)=\mathrm{6} \\ $$$$\:\mathrm{case}\:\mathrm{2}:\:^{\mathrm{6}} {P}_{\mathrm{2}} \:\:\left({r}\:\mathrm{2}{r}\:\mathrm{2}\:{slots}\:{to}\:{be}\:{filled}\:{in}\:\mathrm{6}\right)=\mathrm{30} \\ $$$$\:\:\mathrm{case}\:\mathrm{3}:\:\frac{{P}_{\mathrm{3}} }{\mathrm{3}!}\:\left({r}\:{r}\:{r}\:\mathrm{3}\:{slots}\:{to}\:{be}\:{filled}.\:\mathrm{3}\:{duplicates}\right)=\mathrm{20} \\ $$
Commented by prakash jain last updated on 11/Dec/15
You can also solve this problem with combination formula. 8 steps need to be taken SSSSSSSS (5D, 3R) The problem is choosing 3 steps among 8 where R will be placed remaining position will have D. total number of ways=^8 C_3 =56
$$\mathrm{You}\:\mathrm{can}\:\mathrm{also}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{with} \\ $$$$\mathrm{combination}\:\mathrm{formula}. \\ $$$$\mathrm{8}\:\mathrm{steps}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{taken} \\ $$$$\mathrm{SSSSSSSS}\:\:\:\left(\mathrm{5D},\:\mathrm{3R}\right) \\ $$$$\mathrm{The}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{choosing}\:\mathrm{3}\:\mathrm{steps}\:\mathrm{among} \\ $$$$\mathrm{8}\:\mathrm{where}\:\mathrm{R}\:\mathrm{will}\:\mathrm{be}\:\mathrm{placed}\:\mathrm{remaining} \\ $$$$\mathrm{position}\:\mathrm{will}\:\mathrm{have}\:\mathrm{D}. \\ $$$$\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}=\:^{\mathrm{8}} {C}_{\mathrm{3}} =\mathrm{56} \\ $$
Answered by Rasheed Soomro last updated on 09/Dec/15
Total positions 24.Let every position be specified by coordinates as below: 00 10 20 30 01 11 21 31 02 12 22 32 03 13 23 33 04 14 24 34 05 15 25 35 From 00 3+5=8 moves are possible From mn
$$\mathcal{T}{otal}\:{positions}\:\mathrm{24}.\mathcal{L}{et}\:{every}\:{position} \\ $$$${be}\:{specified}\:\:{by}\:{coordinates}\:{as}\:{below}: \\ $$$$\mathrm{00}\:\:\mathrm{10}\:\:\mathrm{20}\:\:\mathrm{30} \\ $$$$\mathrm{01}\:\:\mathrm{11}\:\:\mathrm{21}\:\:\mathrm{31} \\ $$$$\mathrm{02}\:\:\mathrm{12}\:\:\mathrm{22}\:\:\mathrm{32} \\ $$$$\mathrm{03}\:\:\mathrm{13}\:\:\mathrm{23}\:\:\mathrm{33} \\ $$$$\mathrm{04}\:\:\mathrm{14}\:\:\mathrm{24}\:\:\mathrm{34} \\ $$$$\mathrm{05}\:\:\mathrm{15}\:\:\mathrm{25}\:\:\mathrm{35} \\ $$$$\mathcal{F}{rom}\:\mathrm{00}\:\:\:\mathrm{3}+\mathrm{5}=\mathrm{8}\:{moves}\:\:{are}\:{possible} \\ $$$$\mathcal{F}{rom}\:{mn}\:\: \\ $$
Commented by prakash jain last updated on 09/Dec/15
Consider this 3r+5d 1d+3r+4d 2d+3r+3d 3d+3r+2d 4d+3r+d 4d+r 2r+5d+r 2r+4d+r+d 2r+3d+r+2d 2r+2d+r+3d 2r+d+r+4d and so on.
$$\mathrm{Consider}\:\mathrm{this} \\ $$$$\mathrm{3r}+\mathrm{5d} \\ $$$$\mathrm{1d}+\mathrm{3r}+\mathrm{4d} \\ $$$$\mathrm{2d}+\mathrm{3r}+\mathrm{3d} \\ $$$$\mathrm{3d}+\mathrm{3r}+\mathrm{2d} \\ $$$$\mathrm{4d}+\mathrm{3r}+\mathrm{d} \\ $$$$\mathrm{4d}+\mathrm{r} \\ $$$$\mathrm{2r}+\mathrm{5d}+\mathrm{r} \\ $$$$\mathrm{2r}+\mathrm{4d}+\mathrm{r}+\mathrm{d} \\ $$$$\mathrm{2r}+\mathrm{3d}+\mathrm{r}+\mathrm{2d} \\ $$$$\mathrm{2r}+\mathrm{2d}+\mathrm{r}+\mathrm{3d} \\ $$$$\mathrm{2r}+\mathrm{d}+\mathrm{r}+\mathrm{4d} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}. \\ $$
Commented by Rasheed Soomro last updated on 09/Dec/15
Th^α nK^S .
$$\mathcal{T}{h}^{\alpha} {n}\mathcal{K}^{\mathcal{S}} . \\ $$

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