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**Soroban** Hello, sous!

There are: .$\displaystyle 6^3 = 216$ possible outcomes.

To get a higher total, $\displaystyle B$ must get a 17 or 18.

There are only 4 such outcomes: .$\displaystyle (5,6,6),\;(6,5,6),\;(6,6,5),\;(6,6,6)$

Therefore: .$\displaystyle P(B\text{ gets 17 or 18}) \;=\;\frac{4}{216} \;=\;\frac{1}{54}$

Let: .$\displaystyle V$ - vowel, .$\displaystyle C$ = consonant.

The permutation has the form: .$\displaystyle CCVVCCCV$

The consonants can be placed in: .$\displaystyle 5!\,=\,120$ ways.

The vowels can be placed in: .$\displaystyle 3! \,=\,6$ ways.

Therefore, there are: .$\displaystyle 120\cdot6 \:=\:720$ permutations.

Since there are 28 connecting roads, there must be **8** cities.

I found no neat formula for this problem . . . I had to make a list.

One-step tour

He drives directly from $\displaystyle A$ to $\displaystyle C$: .1 way.

Two-step tour

He drives from $\displaystyle A$ to one of the other 6 cities,

. . then drives to $\displaystyle C$: .$\displaystyle _6P_1 \,=\,6$ ways.

Three-step tour

He drives from $\displaystyle A$ through two of the other 6 cities (in some order)

. . then drives to $\displaystyle C$: .$\displaystyle _6P_2 \,=\,30$ ways.

Four-step tour

He drives from $\displaystyle A$ through three of the other 6 cities (in some order)

. . then drives to $\displaystyle C$: .$\displaystyle _6P_3\,=\,120$ ways.

Five-step tour

He drives from $\displaystyle A$ through four of the other 6 cities (in some order)

. . then drives to $\displaystyle C$: .$\displaystyle _6P_4 \,=\,360$ ways.

Six-step tour

He drives from $\displaystyle A$ through five of the other 6 cities (in some order)

. . then drives to $\displaystyle C$: .$\displaystyle _6P_5 \,=\,720$ ways.

Seven-step tour

He drives from $\displaystyle A$ through all six of the other 6 cities (in some order)

. . then drives to $\displaystyle C$: .$\displaystyle _6P_6 \,=\,720$ ways.

Therefore, there are: .$\displaystyle 1 + 6 + 30 + 120 + 360 + 720 + 720 \;=\;1957$ ways.