Find the maximum distance between A and H if condition is that traveling a city more than once is not allowed.

From solution of question number 26 ,
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths)
Hence , the total number of such paths is 4 .

On the basis of above given diagram , we can say that
The minimum distance between A and H through D is HEDCA and distance is 500 + 500 + 500 + 600 = 2100.
Hence , required answer will be 2100 km.

As per the given above diagram , we can see that
For minimum distance path is HGCA = 400 + 400 + 600 = 1400
Therefore , the minimum distance between H to A without visiting city E is 1400 km .

Number of paths is as follows:
Starting from HGB: HGBA, HGBCA, HGBCDA, HGBCEDA, HGBCEFDA (Total 5 paths)
Starting from HGC: HGCA, HGCBA, HGCDA, HGCEDA, HGCEFDA (Total 5 paths)
Starting from HGE: HGECA, HGECBA, HGECDA, HGEDCA, HGEDCBA, HGEDA, HGEFDCBA, HGEFDCA, HGEFDA (Total 9 paths)
Starting from HEG: HEGBA, HEGCBA, HEGCA, HEGCDA (Total 4 paths)
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths) Starting from HED: HEDCBGA, HEDCBA, HEDCA, HEDA (Total 4 paths)
Starting from HEF: HEFDCGBA, HEFDCBA, HEFDCBA, HEFDCA, HEFDA (Total 5 paths)
Starting from HFE: HFEGBA, HFECGBA, HFECBA, HFECA, HFEDCGBA, HFEDCBA, HFEDCA, HFEDA, (Total 8 paths)
Starting from HFD: HFDEGBA, HFDEGCA, HFDEGBCA, HFDECGBA, HFDECBA, HFDECA, HFDCEBGA, HFDCGBA, HFDCBA, HFDCA, HFDA (Total 11 paths)
Total number of paths is: 5 + 5 + 9 + 4 + 4 + 4 + 5 + 8 + 11 = 55
Hence , the total number of paths from city H to city A is 55 .

From the above given diagram , we can see that
Maximum distance is when path taken is-
A ? D ? C ? B ? G ? E ? F ? H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km
Minimum distance is when path taken is-
A ? C ? G ? H = 600 + 400 + 400 = 1400 km
Hence , Required ratio is 37 : 14.

Map of the town is given as below:
It is given that Neelam took the shortest path hence she has to pass through path EF.
Number of ways to reach E from A is ²⁺²C₂ = ⁴C₂ = 6.
Number of ways to reach B from F is ⁴⁺²C₂ = ⁶C₂ = 15.
Hence total number of ways to reach B from A is 6 x 15 = 90.

From the answer of previous question,
Total number of ways from A to B is 90.
Now we need to calculate total number of ways from B to C that is 13 (on the same logic)
Hence, total number of paths from A to C is 90 x 13 = 1170.

As per the given conditions,
we require surveillance cameras that would cover all roads converging at an intersection. The cost for covering all the roads has to be minimum.
If we put a camera on '1', roads from '7' and '6' to 1 will be covered.
If we put a camera on intersection '2', roads converging from '3' to '2' will be covered.
If we put a camera on '3', roads from '3' to '5', '6' to '5' and '7' to '5' will be covered.
Now we are left with roads from 7 to 4 and from 6 to 4 both of them converge from 4 hence these will be covered by putting a camera on '4'. And then all the roads will be covered and the minimum cost is 2 + 5 + 1 + 4 = 12 lakhs

On the basis of above given question ,we can say that
Similar to the above answer cameras are put on 2, 5, 1 and 4 to minimize the cost.

As per the given above diagram , we can see that
Total number of paths from X to Y is 7.