Help in permutations and combinations ( H2 MATHS )
-
I need help in these questions regarding P and C
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) -
3) 1,1,1,1 - 1choiceOriginally posted by lavastar:I need help in these questions regarding P and C 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)
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 -
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.Originally posted by lavastar:I need help in these questions regarding P and C 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)
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 -
1) there are only 3 S and 3 T.Originally posted by lavastar:I need help in these questions regarding P and C 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)
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 -
4) W,M,C,M,W,M,C,M,WOriginally posted by lavastar:I need help in these questions regarding P and C 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)
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 -
P & C is my weakest... Following on all the statistics I very weak de... can only try my bestOriginally posted by lavastar: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 )
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 -
M M M M M F F F F FOriginally posted by lavastar: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)
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 -
Oops, just realised Gasfene answered everythign liao
-
thanks all
-
No ar. I dont know how to do Q4. Forgot the techniques. Just smoke through.
-
P & C is my weakest topic.
The rest quite easy to understand.
-
Actually my calculations from my understanding for q4 is 288Originally posted by Gasfene:No ar. I dont know how to do Q4. Forgot the techniques. Just smoke through.
Dunno where the extra * 3 come from... -
You may be right. Answer for Q4 seems to be 288.Originally posted by eagle:Actually my calculations from my understanding for q4 is 288
Dunno where the extra * 3 come from...
unless the children must sit on woman's lap
-
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
long time nv touch PnC alr... -
Correct! On the right track...Originally posted by SBS261P: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 long time nv touch PnC alr...
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". -
H2 MATHS seem quite hard now...
-
I've always hated statistics part... Can do, but just hate it...Originally posted by Gasfene:H2 MATHS seem quite hard now...
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