Fiveless|notes



Arranging Letters

| [RJC07/P2/5i] Find the number of different ways in which ‘CONNOISSEUR’ can be arranged, if there are no restrictions. |

Straightforward stuff first. The word has 11 letters in total, 2 Os, 2Ns and 2 Ss.

The total number of arrangements is thus given by

[pic]= 4989600

| [RJC07/P2/5ii] if a vowel must be separated from another with exactly one consonant. |

Take note of cunning questions like this. The word has 6 consonants (C) and 5 vowels (V). At first sight it’s easy to overlook the fact that the word can start or end with a consonant as the question makes no restriction that a consonant has to be separated by a vowel. Thus the only correct arrangements would be:

_V_V_V_V_V_ _ _V_V_V_V_V V_V_V_V_V_ _

The total number of combinations is given by

3[pic] = 10800

|[CJC05/2/3a] Find the number of arrangements of the letters in the world ‘PERSEVERE’ if no 2 ‘E’s are together |

Consider the representation below with the blanks representing where Es can be slotted

_P_R_S_V_R_

Hence, the total number of combinations is given by

|6C4 |× | [pic] |= 900 |

|4 Es to slot into |5 remaining letters to| |

|6 positions |arrange (2 Rs) | |

|[CJC05/2/3b] if each arrangement must start and end with ‘S’ or ‘P’ and no 2 ‘E’s are together |

[P/S]_R_V_R_[S/P]

|2! |× |[pic] |= 900 |

|Number of |3 remaining letters to| |

|positions of P/S |arrange (2 Rs) | |

Since there are only 4 slots to put 4 Es, there’s no need to account for the Es

|[SAJC05/1/2ii] Find the number of different words that can be formed by using all the letters in the word REPTITION, if the first and the|

|last letters are both consonants |

|[pic] |× |2! |× |[pic] |= 100800 |

|Choose 2 consonants from a |Switch them |And arrange the rest | |

|total of 5 (with 2 Ts) |around |(2 Es and 2 Is) | |

|[RJC05/2/1] A school library has 3 books by Dickens, 4 books by Shakespeare and 2 books by Montogomery. Find the number of ways of |

|arranging the books on the shelf if the books written by the same author (i) are indistinguishable; (ii) are distinguishable and must be |

|together; (iii) are distinguishable and all the 3 books by Dickens must be separate. |

This is a letter-arranging question disguised as something else.

|(i) “Find the number of |(ii) You have three bunches of books, within each bunch |(iii) _S_S_S_S_M_M_ |

|arrangements of the letters in |you can shift the books around. |7C3 3! 6! |

|DDDSSSSMM” | |You have 3Ds to slot into 7 positions. Multiply|

|[pic]= 1260 |3! × 3!4!2! = 1728 |by the factorials because the books are |

| |3! to shift the bunches around, and the rest of the |discrete. 3! to arrange the Dickens, 6! to |

| |factorials to move the books around within each bunch. |arrange the rest. |

Arranging People

|[ACJC05/1/10i] Doris, a matchmaker is to organize a dinner for 4 men and 4 women. How many ways can she select these 8 people from a list|

|of 10 men and 10 women? |

Simple stuff – choose 4 people from each of two groups of ten. The number of ways is given by

10C410C4 = 44100

|[ACJC05/1/10iii] (To tell you the gist of the question, one man leaves, and Doris has to split what’s left into 3 groups, each consisting|

|of at least one man and woman) |

So you’re left with three men and four women, which leaves you with no choice but to form two groups of a man and a woman, and one group with one man and two women. The number of ways is given by

|4C2 |× |3! |= 36 |

|Choosing the group | |Moving the 3 men| |

|with 2 women | |around | |

There is no need to multiply the answer by 2! as the pairing of the one-woman groups is already accounted for when you move the three men about, and multiplying by 2! would be double counting. Try listing out the possible combinations if it’s not immediately apparent.

|[JJC05/2/1a] A committee of 5 is to be selected from 6 men and 4 ladies, find the number of committees that contain at least two ladies. |

The trick, of course, is to know when to use the complement

|10C5 |- |4C16C4 |- |6C5 |= 186 |

|All possible |1 woman |0 women | |

|combinations |4 men |5 men | |

|[NYJC05/1/3] A world-class basketball team has ten players and each player is versatile enough to play in any position. If the captain |

|and exactly one of the two youngest players have to be in the starting line-up, find the number of ways of fielding the starting line-up |

|if each line-up must consist of five players. |

Ignore the captain - he has to be in anyway. You are now left with nine people to fill four positions. Of that, only one out of the two youngest can play. That leaves you with seven people to fill three more positions.

|2C1 |× |7C3 |= 70 |

|1 of the 2 youngest | |3 from the | |

|players | |remaining 7 | |

|[TJC05/1/2] (Gist of the question: you have 3 men and 1 lady, and four hotels room available for them to put up in) Find the number of |

|ways in which they can check-in if (i) all of them occupy different rooms, (ii) all 3 men occupy one room and the lady occupies another |

For (i), it’s rather obvious the answer is 4! For (ii), the number of ways is given as follows

|4C2 |× |2! |= 12 |

|Choosing 2 rooms | |Switching the 2 | |

|from 4 | |groups around | |

|[TJC05/1/2iii] …if two men occupy one room, and the other man and lady each occupy different rooms. |

|4C3 × 3! |× |3C2 |= 12 |

|As above (but | |Choosing the 2 men | |

|with 3 rooms) | |occupying 1 room | |

about Circular Tables

|[JJC05/2/1b] If a particular committee is made up of 3 men and 2 ladies, May and June, find the number of ways the committee can be |

|seated in a round table when May and June are not together |

Another complement question. Also remember the formula for circular permutation – (n-1)!

|4! |- |(3! |× |2!) |= 12 |

|All possible | |When M and J are together | |

|combinations | |2! switches their positions | |

|[MI05/2/3ii] 5 males and 4 females are to be seated around the (a?) table. Find the number of ways this can be done if all the girls are |

|seated together. |

All the girls are seating in a group, so effectively, you have six bunches of people to arrange around the table – 5 discrete guys and a bunch of 4 girls.

|5! |× |4! |= 2880 |

|(6-1)! | |Shifting the | |

| | |girls around | |

|[MI05/2/3iii] …if one particular girl must sit between two particular boys |

Now you have seven bunches of people – 3 discrete girls, 3 discrete guys, and a bunch consisting of two guys and a girl.

|6! |× |2! |= 1440 |

|(7-1)! | |Shifting the 2 | |

| | |guys around | |

Other Random Questions

|[RJC05/2/1ii] The digits of the number 3215463 are rearranged. If the first and last digit must be different, find how many such numbers |

|are odd. |

Here is a good time to consider cases.

|Case 1: Ends in 3 – use the complement |Case 2: Ends in 1 or 5 |

|6! |2[pic]= 720 |

|- |2 represents two cases (ends in 1/ends in 5). The second term |

|5! |represents the rearrangement of the remaining terms (2 threes) |

|= 600 | |

| | |

|All | |

| | |

|Starts and ends in 3* | |

| | |

| | |

|5! represents the rearrangement of the remaining 5 letters | |

Total number of combinations = 1320

|[SRJC05/1/5(i)] (Gist: 17 students sit in a circle. Each student has to shake hands with the rest of the class, excluding the two sitting|

|beside him/her in the circle) How many handshakes would have taken place when the game is over? |

This is a rather strange question, not technically a P&C question, but anyway. For questions about handshakes, the total number of handshakes exchanged in a group of n people is given by [pic]. So, in a group of 17 people, the total number of handshakes would be given by 1 + 2 + 3 + … + 15 + 16 = 139.

However, in this case, each person does not shake hands with the two people beside him. For the purposes of this question, imagine a 17-sided polygon with each vertex connected to every other vertex. The number of edges (lines) in this figure is given by 139 – the total number of handshakes without restrictions. Now remove the outer edge of the polygon – 17 lines, which essentially represent the handshakes exchanged between members of the circle sitting beside each other. Thus, the number of handshakes here is given by

139 – 17 = 119

|[SRJC05/1/5(ii)] (Gist: 12 boys and 5 girls have to stand in a row to take a class photo. The boys have to be arranged from shortest to |

|tallest, but do not need to stand together. The girls can stand anywhere they like.) In how many ways can the class be arranged in the |

|row. |

This was a rather strange question as well. How I did this was I imagined I was arranging 17 letters – 12 of which were the same – representing the boys. Then this number was multiplied by two because the arrangement by height can take place both from right to left or left to right.

The total number of ways was thus given by

2[pic]= 1485120

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