-
Best of luck to the class of 2024 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Perm / Comb question - please help (1 Viewer)
- Thread starter kimtuluu
- Start date
Many thanks.
And, by the symmetry property, these two expressions are the same.
My answer for this question was
.Could you please also help with the next 2 questions
1) In how many ways can 12 people be divided into two groups of 6?
2) In how many ways can 12 people be divided into two groups of 6 if two of the people, Jack and Jill, must be in the same group?
Many thanks. Just wonder if you could help explaining how to get to those answers
1) logic works same as here
In how many ways can a group of six people be divided into: 2 equal groups? 2 unequal groups, if there must be at least one person in each group?
In how many ways can a group of six people be divided into: a) two equal groups I have $^6C_3 \times \space ^3C_3 = 20$ So, to choose the first group I have $6$ possibilities of which I am choosi...
2) u fix jack and jill in one group. then 10C4 the other 4. might be wrong here dunno
My advice on these questions is to imagine that you have the people in front of you and have to actually do the task.
So, for the first question, I have 12 people in front of me and I need one group of 5 and one group of 7.
Method 1:
!} \\ &= \cfrac{12!}{5!7!} \\ &= \cfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5 \times 4 \times 3 \times 2 \times 1 \times 7!} \\ &= \cfrac{12 \times 11 \times 2 \times 3 \times 1 \times 1}{1 \times 1 \times 1 \times 1 \times 1 \times 1} \\ &= 792\ \text{possible ways} \end{align*})
Method 2:
!} \\ &= \cfrac{12!}{7!5!} \\ &= \cfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5 \times 4 \times 3 \times 2 \times 1} \\ &= \cfrac{1 \times 11 \times 1 \times 9 \times 8 \times 1}{1 \times 1 \times 1 \times 1 \times 1 \times 1} \\ &= 792\ \text{possible ways} \end{align*})
Method 3:
!}{(\text{Number of "a"'s})!{(\text{Number of "b"'s})!}} \\ &= \cfrac{12!}{5!7!} \\ &= \cfrac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{ 5 \times 4 \times 3 \times 2 \times 1 \times 7!} \\ &= \cfrac{1 \times 11 \times 1 \times 9 \times 8 \times 1}{1 \times 1 \times 1 \times 1 \times 1 \times 1} \\ &= 792\ \text{possible ways} \end{align*})
For the second question, we have 12 people and need two groups of 6. This is very similar to the first question, but with a trick. Using any of the three methods illustrated above, you should get the same result
^2})
However, if we name the 12 people as A, B, C, D, E, F, U, V, W, X, Y, and Z, and place A, B, C, D, E, and F, into group 1 and place U, V, W, X, Y, and Z into group 2, then we have the same outcome as if we place U, V, W, X, Y, and Z into group 1 and A, B, C, D, E, and F, into group 2. In other words, because the groups are the same size and the labels "1" and "2" are arbitrary and not meaningful, we have double counted every possibility. Thus, the actual result is that
^2} = 462\ \text{ways})
Now, if the question had made the groups different then the double counting had not occurred. For example, if we have 12 children who will undertake two activities, and we need one group of six to do activity A then B and another group of six to do activity B then activity A, then the number of ways is 
as the A then B group is different from the B then A group. Similarly, if the two groups were doing the same activity at the same time but one with teacher A and one with teacher B then again there would be 924 ways.
Now, the third question has Jack and Jill amongst the 12 people, and again specifies two groups of six.
Method 1:
!} \\ &= \cfrac{10!}{4!6!} \\ &= \cfrac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!} \\ &= \cfrac{10 \times 3 \times 1 \times 7 \times 1}{1 \times 1 \times 1 \times 1 \times 1} \\ &= 210\ \text{possible ways} \end{align*})
Method 2:
!}{(\text{Number of "J"'s})!{(\text{Number of "N"'s})!}} \\ &= \cfrac{10!}{4!6!} \\ &= \cfrac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!} \\ &= \cfrac{10 \times 3 \times 1 \times 7 \times 1}{1 \times 1 \times 1 \times 1 \times 1} \\ &= 210\ \text{possible ways} \end{align*})
So, for the first question, I have 12 people in front of me and I need one group of 5 and one group of 7.
Method 1:
- I can choose 5 for the first group from the 12 people in 12 choose 5 ways.
- I have seven people left, all of whom need to go into the second group, which can be done in only one way.
- So, the total is
Method 2:
- I can choose 7 for the first group from the 12 people in 12 choose 7 ways.
- I have five people left, all of whom need to go into the second group, which can be done in only one way.
- So, the total is
Method 3:
- I can go through the 12 people, assigning each of them in sequence to either group 1 or group 2.
- What results is a arrangement of 12 items, five labelled as '1' and seven labelled as '2', which is identical (in combinatorial terms) to a "word" made up of 12 letters, five of which are "a"'s and seven of which are "b"'s.
- So, the total is
However, if we name the 12 people as A, B, C, D, E, F, U, V, W, X, Y, and Z, and place A, B, C, D, E, and F, into group 1 and place U, V, W, X, Y, and Z into group 2, then we have the same outcome as if we place U, V, W, X, Y, and Z into group 1 and A, B, C, D, E, and F, into group 2. In other words, because the groups are the same size and the labels "1" and "2" are arbitrary and not meaningful, we have double counted every possibility. Thus, the actual result is that
Now, the third question has Jack and Jill amongst the 12 people, and again specifies two groups of six.
Method 1:
- We can now designate a Jack and Jill group and place them into it in 1 way.
- The remaining four people in this group need to be selected from the 10 left, and so in 10 choose 4 ways.
- We are then left with people to go in the second group, which can be done in only one way.
- The groups are now not interchangeable because one has Jack and Jill and one does not.
- So, the total is
Method 2:
- I can place Jack and Jill in the "J" group in 1 way.
- I can go through the 10 remaining people, assigning each of them as "J" (the Jack and Jill group) or "N", the non-Jack-and-Jill group.
- For two groups of six in total, this group of 10 people needs to have four people labelled as "J" and six labelled as "N". This is identical (in combinatorial terms) to a "word" made up of 10 letters, four of which are "J"'s and six of which are "N"'s.
- So, the total is
Many thanks for your helpMy advice on these questions is to imagine that you have the people in front of you and have to actually do the task.
So, for the first question, I have 12 people in front of me and I need one group of 5 and one group of 7.
Method 1:
- I can choose 5 for the first group from the 12 people in 12 choose 5 ways.
- I have seven people left, all of whom need to go into the second group, which can be done in only one way.
- So, the total is
Method 2:
- I can choose 7 for the first group from the 12 people in 12 choose 7 ways.
- I have five people left, all of whom need to go into the second group, which can be done in only one way.
- So, the total is
Method 3:
- I can go through the 12 people, assigning each of them in sequence to either group 1 or group 2.
- What results is a arrangement of 12 items, five labelled as '1' and seven labelled as '2', which is identical (in combinatorial terms) to a "word" made up of 12 letters, five of which are "a"'s and seven of which are "b"'s.
- So, the total is
For the second question, we have 12 people and need two groups of 6. This is very similar to the first question, but with a trick. Using any of the three methods illustrated above, you should get the same result
However, if we name the 12 people as A, B, C, D, E, F, U, V, W, X, Y, and Z, and place A, B, C, D, E, and F, into group 1 and place U, V, W, X, Y, and Z into group 2, then we have the same outcome as if we place U, V, W, X, Y, and Z into group 1 and A, B, C, D, E, and F, into group 2. In other words, because the groups are the same size and the labels "1" and "2" are arbitrary and not meaningful, we have double counted every possibility. Thus, the actual result is that
Now, if the question had made the groups different then the double counting had not occurred. For example, if we have 12 children who will undertake two activities, and we need one group of six to do activity A then B and another group of six to do activity B then activity A, then the number of ways is
Now, the third question has Jack and Jill amongst the 12 people, and again specifies two groups of six.
Method 1:
- We can now designate a Jack and Jill group and place them into it in 1 way.
- The remaining four people in this group need to be selected from the 10 left, and so in 10 choose 4 ways.
- We are then left with people to go in the second group, which can be done in only one way.
- The groups are now not interchangeable because one has Jack and Jill and one does not.
- So, the total is
Method 2:
- I can place Jack and Jill in the "J" group in 1 way.
- I can go through the 10 remaining people, assigning each of them as "J" (the Jack and Jill group) or "N", the non-Jack-and-Jill group.
- For two groups of six in total, this group of 10 people needs to have four people labelled as "J" and six labelled as "N". This is identical (in combinatorial terms) to a "word" made up of 10 letters, four of which are "J"'s and six of which are "N"'s.
- So, the total is