Yes.

When the bug is on each point, it can only move up (U) or right (R)

Number of ways to choose to move from 1 point to the next = 2 (either U or R)

From A to C, it has to pass 4 different points (including C)

What we notice is that only 2 upward movements and 2 rightward movements are needed to move to C.

Eg.
R R U U
U U R R
R U R U
U R U R
R U U R
U R R U


These give different unique paths to C.

Number of different ways to get to C
= 6

Likewise, from C to B, we only need 1 upward step and 2 rightward steps.

Eg.

U R R
R R U
U R R

number of ways = 3

Total number of ways

= Number of ways to get from A to C × number of ways to get from C to B

= 6 × 3
= 18

The formula for such questions involves permutations.


From A to C, the 6 ways can be calculated like this

4! ÷ 2! ÷ 2!

= 4 × 3 × 2 × 1 ÷ (2×1) ÷ (2×1)

= 24 ÷ 2 ÷ 2

= 6


! stands for factorial. It means the product of consecutive whole numbers from 1 all yhe way up to the number written next to the !

Eg. 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1


What we are doing is that we a finding how many unique possible permutations of two R's and two U's there are.

_ _ _ _

That 4! stands for :

4 ways for first letter
3 ways for second letter
2 ways for third letter
1 way for fourth letter.


But since there are 2 identical Us and 2 identical Rs, we will get repeated cases.


Lets first look at a hypothetical case where that the two Us are different (U₁ and U₂)


Then, for the permutation U₁U₂ R R would be different from U₂ U₁ R R. These would be two different ways.

Likewise for R U₁ U₂ R and R U₂ U₁ R. They would be different.



But for our actual question, the two Us (and Rs) are identical.


Eg. in U U R R , swapping the 2 Us makes no difference but if we used the above method, we would have counted it as 2 unique ways. That is double counting.


So for those letters with more than one copy, we have to divide by the number of permutations they can generate to get rid of repeated cases.

That's why we have the ÷ 2! ÷ 2!

Similarly, from C to B,

we have one U and two R

Number of ways

= 3! ÷ 2! (to remove repeated cases for the Rs)

= 3 × 2 × 1 ÷ (2 × 1)

= 3 × 2 × 1 ÷ 2 ÷ 1

= 3

There is another similar method which involves labelling of the number of ways for each point.

It can be found in Terry Chew's book for P5 (Solving Exam Type Problem)

Ummm...... sorry I am a little confused is there any simple explanation

This permutation and combination is usually learnt at Sec 4/ A level. So that's why more difficult to grasp.

Oh so does that mean that there is no other simple method ?

Have. Later I send directly