Help in permutations and combinations ( H2 MATHS )

lavastar SG

I need help in these questions regarding P and C Embarassed i included the answers in brackets beside it.. Thx for those who can help

1) Four letters from the word STATISTICS are taken to form a code-word. How many of these code words contain

i) only S and T (14)
ii) Neither T nor C ( 38 )

2) Five married couples are to be seated in a row of 10 adjacent seats for a concert. In how many ways can they be seated if

i) all the men sit in a group while all the women sit together given that no husband sit next to his wife. (23040)

3) How many different 4-digit numbers can be formed using the digits 1,1,1,1,2,3,4,8 if

i) There are no restrictions (209)

4) Four men, three women and two children are to be seated in a row under the following restrictions : each child must be seated between 2 men and no women must sit next to each other. Find the number of possible seating arrangements. (864)

Gasfene SG

3) 1,1,1,1 - 1choice
1,1,1,2-4 permutations
1,1,1,3-4 permutations
1,1,1,4-4 permutations
1,1,1,8-4 permutations

1,1,2,3-12 permutations (4P2)
1,1,2,4-12
1,1,2,8-12
1,1,3,4-12
1,1,3,8-12
1,1,4,8-12

1,2,3,4-24permutations (4P 3)
1,2,3,8-24
1,2,4,8-24
1,3,4,8-24
2,3,4,8-24

altogether=209

Gasfene SG

2) First in order for men and women to sit in their own gender group, one gender group must be on the left, the other on the right. It can be any gender on each side so *2.

all the man sit in a group while all the woman sit together (no other restriction)
2*5!*5!=28800

Suppose one couple must sit togther in the middle.
2*4!*4!=1152

5 couples: 1152*5=5760

total with restriction: 28800-5760=23040

Gasfene SG

1) there are only 3 S and 3 T.

to form a 4 letter code word: you can either have 3S, 1T or 2S, 2T and 3T, 1S
first have 4 permutations, second one have 6permutations (4choose 2), third have 4 permutations.

altogether 4+6+4=14

ii)Saisis
3s, 2i, 1a
code word: either 3s,1i, or 3s, 1a or 2s 2i, or 2 s 1i 1a, or 1s, 2i,1a
4+4+6+12+12=38

Gasfene SG

4) W,M,C,M,W,M,C,M,W

this is the only combination possible since no 2 women must sit togehter and each child must sit with 2 guys.

treat each M,C,M as 1 unit called U

W,U,W,U,W

arrange the 3 women in 3! ways
arrange the 2 units in 2! ways

fuck i dunno. anyhow working to get answer. EDIT: 3!*2!*4!=288
6 is obtained from 4choose2

eagle SG

P & C is my weakest... Following on all the statistics I very weak de... can only try my best Embarassed

i) 3 letters of S and 3 letters of T

SSST type ~ 4!/3! = 4 words
SSTT type ~ 4! / (2! * 2!) = 6 words
STTT type ~ 4!/3! = 4 words

add up = 14 words

or you could view it as for each of the letter, you have 2 choices between S or T
which means,
2 * 2 * 2 * 2 words = 16 words
But this includes SSSS and TTTT, so 16 - 2 = 14 words

ii)
Remaining: S S S A I I

Num of S
= 3: then either A or I (2 * 4!/3!)
= 2: then either with AI (4!/2!) or II (4!/{2!*2!})
= 1: then AII (4!/2!)

Sum up = 38 choices

eagle SG

M M M M M F F F F F

like that combi is 5! * 5! * 2 = 28800 ways

M M M M couple F F F F

like that combi is 4! * 4! * 2 * 5 (couples choice) = 5760 ways

Hence, answer is 28800 - 5760 = 23040 ways

eagle SG

Oops, just realised Gasfene answered everythign liao

lavastar SG

thanks all Smile

Gasfene SG

No ar. I dont know how to do Q4. Forgot the techniques. Just smoke through.

SBS3625D SG

P & C is my weakest topic.

The rest quite easy to understand. Razz

eagle SG

Actually my calculations from my understanding for q4 is 288

Dunno where the extra * 3 come from...

Gasfene SG

You may be right. Answer for Q4 seems to be 288.

unless the children must sit on woman's lap Twisted Evil

SBS261P SG

i think most of us got 288 because we only consider:

W (M C M) W (M C M) W

but we forget to consider

W (M C M C M) W (M) W

for the first case ---- number of ways = 3! x 4C2 x 2! x 2 x 2! = 288
[permute de 3 women first (3!), den choose 2 men out of 4 (4C2), den permute de men in 1 set (2!), den choose 1 child out of 2 (2), den for the other set u can permute de men also (2!)]

for the second case ---- number of ways = 3! x 4C3 x 3! x 2! x 2 = 576
[permute de 3 women first (3!), den choose 3 men out of 4 (4C3), den permute de men in 1 set (3!), den permute de 2 children (2!), den swap de 2 bracketed sets (2)

den add both together: total number of ways = 576 + 288 = 864

Very Happy long time nv touch PnC alr...

eagle SG

Correct! On the right track...

I would say

W (M C M) W (M C M) W

W (M C M C M) W (M) W

and

W (M) W (M C M C M) W

each of it is 288

so 288 * 3 = 864

I do by how many choices there are for each "box".

Gasfene SG

H2 MATHS seem quite hard now…

eagle SG

I've always hated statistics part... Can do, but just hate it...

For C maths last time, I chose all mechanics and 0 statistics
For F maths, do 7 questions from choices of 5 mechanics and 5 stats, I always do 5 mechanics and 2 stats Embarassed