P&C

wishboy SG

Hi, I stumbled upon this question in my tutorial. It is taken from MJC/2009/P2/8.

Find the number of different arrangements of the name "BENNETT TAN"

(i) beginning with B and ending with N, Ans: 1680
(ii) with all the T's separated and all the N's separated. Ans: 12120

I got the answer for (i) by

8!/(2!2!3!) = 1680

but I'm not sure how to do (ii), how do I approach this question?

Thanks in advance! :D

wee_ws SG

Hi,

The steps are:

1. Arrange A, B, E, E.

2. Slot N, N, N in between A, B, E, E.

3. Slot T, T, T in between the letters in step 2.

The correct answer is 6720.

Do take note that many P&C questions have the following problem-solving format:

- Arrange the others.
- Slot things in between so that they are separated.

In the question we just saw, the process of slotting is carried out twice. Thanks!

Cheers,
Wen Shih

wishboy SG

Hi,

The steps are:

1. Arrange A, B, E, E.

2. Slot N, N, N in between A, B, E, E.

3. Slot T, T, T in between the letters in step 2.

The correct answer is 6720.

Do take note that many P&C questions have the following problem-solving format:

- Arrange the others.
- Slot things in between so that they are separated.

In the question we just saw, the process of slotting is carried out twice. Thanks!

Cheers,
Wen Shih

Yeah I thought about that initially, but couldn't get the answer provided.

Consider this case,

ABEETNTNTN (and other similar cases such as ABNTNTNTEE or ANTNTBENTE

This is not attainable if I use the slotting in twice method, since I would be slotting more than 1 N into a slot.

wee_ws SG

Hi,

You have a great point that I did not consider :P

Back to the drawing board...

Cheers,
Wen Shih

SBS261P SG

Mad Hat SG

The question is nasty. I can't find a simple way. Here's an ugly way. Anyone got a better way do share. Students should not be worried about this sort of question, which is sadistic!

abee can be permuted in 12 ways

Now add three Ts
Case 1 ................. a b e e TTT ................. 5 ways
Case 2 ................. a b e T e TT................. ⁵C₂ x 2 ways
Case 3 ................. a b T e T e T................. ⁵C₃ ways

Now add three Ns, the red ones being forced
Case 1 ................. N a b e e TNTNT ................. 6 ways
Case 2 ................. N a N b e T e TNT ................. ⁷C₂ ways
Case 3 ................. N a N b N T e T e T................. ⁸C₃ ways

Total
(12 x 5 x 6) + (12 x ⁵C₂ x 2 x ⁷C₂) + (12 x ⁵C₃ x ⁸C₃) = 12,120

wishboy SG

Wow thanks!

Wondering if there is any simpler method to do this question...

eagle SG

Yeah I thought about that initially, but couldn't get the answer provided.

Consider this case,

ABEETNTNTN (and other similar cases such as ABNTNTNTEE or ANTNTBENTE

This is not attainable if I use the slotting in twice method, since I would be slotting more than 1 N into a slot.

You are good! The solution provided by MJC itself says 6720.

Looking at the solution, I guess this part wasn't on the examiner's mind when he/she set the paper.

I think Mad Hat's method is very good and logical and easy to understand.

wee_ws SG

Hi Mad Hat,

Thanks for sharing your solution which is a very good method. I agree with you that this question is highly difficult and is thus not suitable for A-level.

Cheers,
Wen Shih

Mad Hat SG

My first answer was 6720 also