A Level Maths P&C Qns
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Hello, I've encountered some problems while doing some P&C qns and I would much appreciate if any maths guru can help me out..
1. A painter is given the job of painting the doors of 5 adjacent bedrooms using 5 colours: red, blue, yellow, green and orange.
If the owner of the house insists that the doors should be painted with at most 2 colours, find the number of ways which the painter could have the job done. [4]
Ans: 305
2.There're 12 boys and 8 girls. 2 groups, with 3 people in each group, are to be formed. Find the number of different ways in which this can be done if:
(i) there're no restrictions
(ii) members within each group are of the same gender
Ans: (i) 387600; (ii) 21840
3. The teacher would like to take a picture of the class consisting of 12 boys and 5 girls standing in a row. The teacher wants the boys to be arranged in order from the shortest to the tallest but they do not need to stand together. The girls are free to stand anywhere. How many ways can the class be arranged in a row? [3]
Ans: 742560
4: In a group of 3 men and 4 women, how many ways can 1 split them into 3 groups, if each group is to consist of at least 1 man and at least 1 woman? [2]
Ans: 28
5: There are 17 students in a class seated in a circle. The students are to find out the names of the persons seated next to them. Next, each student is required to go round and exchange handshakes with the rest of the class, excluding the 2 students seated next to him or her. How many handshakes would have taken place when the game is over?
- I can work out the answer (119) by manually summing up all the cases: 14+14+13+12+...., but I would like to understand the basis behind the answer: (14x17)÷2!=119.. How was the 2! derived?
My workings:
For 1,
5C2 [Choose 2 colours 1st] x 2^5 [Each of the 5 rooms can have 2 colours] = 320
320 - (5C2-5) [minus the repetition of colours chosen] = 315
Where did I overcount/ What's wrong with my subtraction?
For 2,
(i) 20C3x17C3=775200
(Why do I need to ÷2?)
(ii) Ans: 12C3x8C3 (why no need ÷2) + 8C3x5C3÷2 (why need to ÷2) + 12C3x9C3÷2 (why need to ÷ 2) = 21840
3: ???
4: 3C1X4C1+2C1X3C1+1C1X2C2 = 19 +2 (since the extra woman can be in 1st/2nd group)= 21
Thanks a lot.. :) Doing P&C qns can be rather addictive so I did quite a handful haha..
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1) Not sure how to get the answer. You must realise that each door can also be painted with a single colour, rather than 2 colours.
All I can think of is for each door, there are 5 ways to paint it with 1 colour, and 5C2 = 10 ways of painting it with 2 colours. So for 5 doors just multiply by 5, which results in 75 ways.2i)
20C3 * 17C3 / 2 = 387 600
divide by 2 because to account for double counting (there's a possibility of choosing group A then B, and choosing B then A, both resulting the same 2 groups of 3 people. this is because they are chosen from the same group of people.)2ii)
3 cases:
2 groups of boys: 12C3 * 9C3 / 2
2 groups of girls: 8C3 * 5C3 / 2
1 group of boys and 1 group of girls: 12C3*8C3 (there is no need to divide by 2 in this case as girls and boys are chosen respectively from their different groups)Add them up and you get 21840.
3)
Did quite a long method, a lot of cases...
There are 12 boys, so there are 13 spaces for the girls to stand in.
(ie. _B_B_B_B_B_B_B_B_B_B_B_B_)
Each slot can contain more than 1 girl.1. All 5 in 1 slot: 13C1 * 5!
2. 4 girls in 1 slot, 1 girl in another slot: 13C2 * 2! * 5!
3. 3 , 2 : 13C2 * 2! * 5!
4. 3 , 1 , 1 : 13C3 * 3!/2! * 5!
5. 2 , 2 , 1 : 13C3 * 3!/2! * 5!
6. 2 , 1 , 1 , 1 : 13C4 * 4!/3! * 5!
7. 1 , 1 , 1 , 1 , 1 : 13C5 * 5!Add each case up and you'll get 742560.
4) No idea. I keep getting 36 though.
5) No idea why divide by 2!
Hope that helps.
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Originally posted by gohby:
Hello, I've encountered some problems while doing some P&C qns and I would much appreciate if any maths guru can help me out..
1. A painter is given the job of painting the doors of 5 adjacent bedrooms using 5 colours: red, blue, yellow, green and orange.
If the owner of the house insists that the doors should be painted with at most 2 colours, find the number of ways which the painter could have the job done. [4]
Ans: 305
2.There're 12 boys and 8 girls. 2 groups, with 3 people in each group, are to be formed. Find the number of different ways in which this can be done if:
(i) there're no restrictions
(ii) members within each group are of the same gender
Ans: (i) 387600; (ii) 21840
3. The teacher would like to take a picture of the class consisting of 12 boys and 5 girls standing in a row. The teacher wants the boys to be arranged in order from the shortest to the tallest but they do not need to stand together. The girls are free to stand anywhere. How many ways can the class be arranged in a row? [3]
Ans: 742560
4: In a group of 3 men and 4 women, how many ways can 1 split them into 3 groups, if each group is to consist of at least 1 man and at least 1 woman? [2]
Ans: 28
5: There are 17 students in a class seated in a circle. The students are to find out the names of the persons seated next to them. Next, each student is required to go round and exchange handshakes with the rest of the class, excluding the 2 students seated next to him or her. How many handshakes would have taken place when the game is over?
- I can work out the answer (119) by manually summing up all the cases: 14+14+13+12+...., but I would like to understand the basis behind the answer: (14x17)÷2!=119.. How was the 2! derived?
My workings:
For 1,
5C2 [Choose 2 colours 1st] x 2^5 [Each of the 5 rooms can have 2 colours] = 320
320 - (5C2-5) [minus the repetition of colours chosen] = 315
Where did I overcount/ What's wrong with my subtraction?
For 2,
(i) 20C3x17C3=775200
(Why do I need to ÷2?)
(ii) Ans: 12C3x8C3 (why no need ÷2) + 8C3x5C3÷2 (why need to ÷2) + 12C3x9C3÷2 (why need to ÷ 2) = 21840
3: ???
4: 3C1X4C1+2C1X3C1+1C1X2C2 = 19 +2 (since the extra woman can be in 1st/2nd group)= 21
Thanks a lot.. :) Doing P&C qns can be rather addictive so I did quite a handful haha..
For question 1, let us consider it from another angle.
If we use 2 paints, there is (2^5 - 2) ways to paint the doors.
I believe you understand why we need the 2^5, but not why we need the -2.
Note that 2 of the combinations will result in all the doors having the same colour of one particular paint. However, we are concerned with the number of combinations possible with 2 different paint colour.
Moving on, there are 5C2 ways to select two different paints from the 5 type of paint available.
Hence, in total there are (5C2) x (2^5 - 2) = 300 ways for us to use 2 different paints to paint the doors.
On the otherhand, there are only 5 ways if we are to use exactly only 1 paint, since there are only 5 types of paint.
Hence, we have a total of 300 + 5 = 305 ways in all.
For 2(i), note that the groups are not labelled. Thus, both groups are interchangable.
For example, the first group having A,B,C in it and the second group having D,E,F in it is the same as the first group having D,E,F in it and the second group having A,B,C in it, since we are not labelling the group.
Hence, we need to divide by 2 to avoid repeating the same groups, as we are not concern about the order here.
For 2(ii), the reason why we need to divide by 2 for the cases where both groups are all girls or all boys is because we want to avoid any repeating, as I have explain in the 2(i).
However, for the one group boy, one group girl case, note that we are choosing from entirely different sets of people. In the previous cases, we are recycling the same set of people - E.g. After getting 3 boys from the set of 12 boys, we get another 3 from the remaining 9 boys.
In the one group girl one group boy case, we are taking 3 boys from a set of 12 boys, and 3 girls from a set of unrelated 8 girls.
For 3, I can say it is really an unfriendly question.
First of all, we have 12 boys who are standing in a manner which cannot be changed.
Now, imagine them to be 12 cones. You can put something between each cone, as well as at each end of the cones. In short, we have 13 places to put the girls in.
Here is the nasty part which may take a while for you to understand. I am sorry but I am not thinking or typing very well at this time in the morning.
We can have:
-all five girls standing apart, separated by the boys
E.g. BBBBGBBBGBBGBGBGB
-two girls sanding together, all three other girls separated by the boys
-two girls standing together, another two standing together, and the last standing apart
- three girls standing together, the other two standing away from each other.
- three girls standing together, the two standing together
- four girls standing together, one girl standing apart
- five girl standing together
Just to make things worse, when you have:
- one group of 2 girls, 3 groups of 1 girl, there are 4 ways to arrange the groups
- two groups of 2 girls, 1 group of 1 girl, there are 3 ways to arrange the groups
- one group of 3 girls, 2 groups of 1 girl, there are 3 ways to arrange the groups
- one group of 3 girls, one group of 2 girls, there are 2 ways to arrange the group
- one group of 4 girls, one group of 1 girl, there are 2 ways to arrange the group
All in all, we have a total of:
(13C5 x 5!) + (13C4 x 5! x 4) + (13C3 x 5! x 3) + (13C3 x 5! x 3) + (13C2 x 5! x 2) + (13C2 x 5! x 2) + (13C1 x 5!) = 742560
For 4), I cannot get the answer, so you will have to wait for my boss or eagle to come here.
For 5), There are 17 people. Each person will shake the hand of 14 people. Assume that everybody actually does that on their own initiative, we will have 17 x 14.
However, this means that everyone is shaking the hand of those who have previously offered to shake hand with them. Hence, there is a repeat of hand-shaking.
Thus, we have (17 x 14)/2 = 119
Where does the factorial comes from is something which I have no idea about though (or is unable to recall due to fatigue).
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as all the other questions are being answered, i will answer 4 only.
FIRSTLY, you copied the wrong answer. The right answer is 36.
How do I know this? I know where all your questions are from. 28 is for part (ii), the question about number of handshakes of the 8 people before 1 left in part (iii).
How to get 36?
Firstly, there's 3 groups, each group will have at least 1 man and 1 woman, So to make things simpler, i will use the man to mark the group and the woman can be chosen into the "man's group".
So here we have
Man A Man B Man C
W1W2 W3 W4.
Number of ways the women (W1,W2,W3,W4) can be divided into the three groups is
3 X 4!/2! = 36. -
Originally posted by Stgrfg:
as all the other questions are being answered, i will answer 4 only.
FIRSTLY, you copied the wrong answer. The right answer is 36.
How do I know this? I know where all your questions are from. 28 is for part (ii), the question about number of handshakes of the 8 people before 1 left in part (iii).
How to get 36?
Firstly, there's 3 groups, each group will have at least 1 man and 1 woman, So to make things simpler, i will use the man to mark the group and the woman can be chosen into the "man's group".
So here we have
Man A Man B Man C
W1W2 W3 W4.
Number of ways the women (W1,W2,W3,W4) can be divided into the three groups is
3 X 4!/2! = 36.I thought that I was actually tired last night (this morning?) so I came out with the wrong answer for Q4.
I am glad you are here to clarify with the TS as I was worried I turned a bit rusty.
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Originally posted by wishboy:
1) Not sure how to get the answer. You must realise that each door can also be painted with a single colour, rather than 2 colours.
All I can think of is for each door, there are 5 ways to paint it with 1 colour, and 5C2 = 10 ways of painting it with 2 colours. So for 5 doors just multiply by 5, which results in 75 ways.2i)
20C3 * 17C3 / 2 = 387 600
divide by 2 because to account for double counting (there's a possibility of choosing group A then B, and choosing B then A, both resulting the same 2 groups of 3 people. this is because they are chosen from the same group of people.)2ii)
3 cases:
2 groups of boys: 12C3 * 9C3 / 2
2 groups of girls: 8C3 * 5C3 / 2
1 group of boys and 1 group of girls: 12C3*8C3 (there is no need to divide by 2 in this case as girls and boys are chosen respectively from their different groups)Add them up and you get 21840.
3)
Did quite a long method, a lot of cases...
There are 12 boys, so there are 13 spaces for the girls to stand in.
(ie. _B_B_B_B_B_B_B_B_B_B_B_B_)
Each slot can contain more than 1 girl.1. All 5 in 1 slot: 13C1 * 5!
2. 4 girls in 1 slot, 1 girl in another slot: 13C2 * 2! * 5!
3. 3 , 2 : 13C2 * 2! * 5!
4. 3 , 1 , 1 : 13C3 * 3!/2! * 5!
5. 2 , 2 , 1 : 13C3 * 3!/2! * 5!
6. 2 , 1 , 1 , 1 : 13C4 * 4!/3! * 5!
7. 1 , 1 , 1 , 1 , 1 : 13C5 * 5!Add each case up and you'll get 742560.
4) No idea. I keep getting 36 though.
5) No idea why divide by 2!
Hope that helps.
Hello.. Thank you very much for your time and your efforts - it was certainly helpful.. :) Sorry for qn 4 though, my bad for typing the wrong answer.. haha
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Originally posted by Forbiddensinner:
For question 1, let us consider it from another angle.
.....
Hi Forbiddensinner, thanks a lot for your elaborate and comprehensive explanation (in the wee hours somemore).. really appreciate it!
Sorry for the confusion / frustration that may have arisen from q4, my bad for erroneously typing the wrong answer.. was a little zonked out too when i transcribed the questions.. -
Originally posted by Stgrfg:
as all the other questions are being answered, i will answer 4 only.
FIRSTLY, you copied the wrong answer. The right answer is 36.
How do I know this? I know where all your questions are from. 28 is for part (ii), the question about number of handshakes of the 8 people before 1 left in part (iii).
How to get 36?
Firstly, there's 3 groups, each group will have at least 1 man and 1 woman, So to make things simpler, i will use the man to mark the group and the woman can be chosen into the "man's group".
So here we have
Man A Man B Man C
W1W2 W3 W4.
Number of ways the women (W1,W2,W3,W4) can be divided into the three groups is
3 X 4!/2! = 36.Hello.. Thanks for your explanation, and incisively pointing out my error in answer! I can understand 3 (the duo can be under Man A/B/C grp) but I'm having a little difficulty understanding 4! (permutating all the women), then dividing by 2 (since W1W2=W2W1). but this doesn't resemble a repetition case with repeating letters/stuff. Just a side note, why is it 2! instead of 2? In this qn, it will not affect the answer but supposing if the qn were to be tweaked (with 5W and each group can have either 1 or 3W) the answer will be different.
Cheers..
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Originally posted by gohby:
Hello.. Thanks for your explanation, and incisively pointing out my error in answer! I can understand 3 (the duo can be under Man A/B/C grp) but I'm having a little difficulty understanding 4! (permutating all the women), then dividing by 2 (since W1W2=W2W1). but this doesn't resemble a repetition case with repeating letters/stuff. Just a side note, why is it 2! instead of 2? In this qn, it will not affect the answer but supposing if the qn were to be tweaked (with 5W and each group can have either 1 or 3W) the answer will be different.
Cheers..
No worries about the typo.
Another slightly longer but easier way to look at this question will be this:
There are 3 groups with 1 man in each group.
Let the men be A, B and C.
We have 4 women, 3 groups, and each group must have at least 1 women.
So there MUST be 2 groups with 2 women and 1 group with 1 women.
Suppose A's group has 2 women.
We have: 4C2 x 2C1 x 1 = 12
Suppose B's group has 2 women.
We have: 4C1 x 3C2 x 1 = 12
Suppose C's group has 2 women.
We have: 4C1 x 3C1 x 2C2 = 12
Hence, we have 12 + 12 + 12 = 36 possible combinations in all.
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For question 3, I was asking my teacher about it. He gave me a very straightforward answer.
Total 17 people - 17!
12 boys are fixed - divide by 12!
so answer is just 17!/12! = 742560
Another way of thinking: Let each boy be a stick (eg. | ) , then let each girl be x.
Therefore we are supposed to rearrange these 12 "identical" sticks and 5 "distinct" girls in a single row.||||||||||||xxxxx
Since the sticks are "identical", no. of ways = 17!/12! = 742560