Question

There are two routes joining a city A to B and three routes joining B to another city C. In how many ways can a person perform a journey from A to C ?​

Answer 1

Answer:

I have solved ur question in paper u can see it

Step-by-step explanation:

i hope it helps u mate!!!!!

Answer 2

Answer:

Initially from A to D through B and C

We have = 4 x 3 x 2 = 24 ways

While returning from D to A through B and C we cannot choose the previous paths

So from D to C instead of 2 ways we have one

From C to B instead of 3 ways we have 2

And from B to A instead of 4 we have 3 ways

So number of ways we have from D to A will be = (4–1) x (3–1) x (2–1) = 3x2x1 = 6

Now total round trip ways become = 24 x 6 = 144

Taking example of above image we have

From A to D the following ways

aeh, aei, afh, afi, agh, agi

beh, bei, bfh, bfi, bgh, bgi

ceh, cei, cfh, cfi, cgh, cgi

deh, dei, dfh, dfi, dgh, dgi. (24)

Now suppose we take “aeh” as route so all possibilities on track a,e and h will be eliminated

Now on returning we left with

ifb, igb, ifc, igc, ifd, igd. (6)

Combining them we have 6 x 24 = 144 ways. Example for aeh we have round trip options

aeh-ifb, aeh-igb, aeh-ifc, aeh-igc, aeh-ifd, aeh-igd (6)

Similarly

6 possibilities each for next 23 A to D options

Hope it will help

Step-by-step explanation: