Perms and Combs (1 Viewer)

Gtsh

New Member
I need help with these questions, I'm so confused

1. The diagram shows a 6 × 4 grid. The aim is to walk from the point A in the top left-hand corner to the point B in the bottom right-hand corner by walking along the black lines either downwards or to the right. A single move is defined as walking along one side of a single small square, thus it takes you ten moves to get from A to B.
Find how many different routes are possible:
I) without restriction,
ii) if you must pass through C

2. Twelve points are arranged in order around a circle.
b) In how many pairs of such triangles are the vertices of the two triangles distinct?
c) In how many such pairs will the triangles: i not overlap, ii overlap?

3. Find how many arrangements of the letters of the word TRANSITION are possible if
a) an N occupies the first but not the last position
b) the letter N is not at either end,
c) the vowels are together.

4. Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.
a)How many such numbers are there?
b) How many of them are odd?
c) How many of them are divisible by 5?
d) How many of them are divisible by 3?

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yanujw

Active Member

i) You must move down a total of 6 times, and right a total of 4 times for a total of 10 moves. The order in which 6 down moves can be placed into a seqeuence of 10 moves is $\bg_white \binom{10}{6} =210$. You may also notice you get the same result if you think of arranging 4 right moves into 10, which is $\bg_white \binom{10}{4} = 210$

ii) This is split into two seperate 'journeys' where you first move from A to C, then C to B. A to C is two moves down and two moves right, creating a total of $\bg_white \binom{4}{2}$ ways for that path. Then from C to B there are $\bg_white \binom{6}{2}$ ways. The total ways is the product of these two, which is 90.

5uckerberg

Active Member
1. The diagram shows a 6 × 4 grid. The aim is to walk from the point A in the top left-hand corner to the point B in the bottom right-hand corner by walking along the black lines either downwards or to the right. A single move is defined as walking along one side of a single small square, thus it takes you ten moves to get from A to B.
Find how many different routes are possible:
I) without restriction,
ii) if you must pass through C
Where is the diagram?

Gtsh

New Member
View attachment 34648
i) You must move down a total of 6 times, and right a total of 4 times for a total of 10 moves. The order in which 6 down moves can be placed into a seqeuence of 10 moves is $\bg_white \binom{10}{6} =210$. You may also notice you get the same result if you think of arranging 4 right moves into 10, which is $\bg_white \binom{10}{4} = 210$

ii) This is split into two seperate 'journeys' where you first move from A to C, then C to B. A to C is two moves down and two moves right, creating a total of $\bg_white \binom{4}{2}$ ways for that path. Then from C to B there are $\bg_white \binom{6}{2}$ ways. The total ways is the product of these two, which is 90.
Thank you so much, that makes sense now!

jimmysmith560

Le Phénix Trilingue
Moderator
Question 2:

Part (b):

$image=https://latex.codecogs.com/png.image?\dpi{110}%20\frac{12C3\times%20\:9C3}{2}=9240&hash=42e6caa90339dc0eecd031e263e3ac23$

You have to divide by 2 because you can choose the same two triangles in two orders, so you have overcounted by a factor of 2.

Part (c):

i) You arbitrarily select an original point, and there are then 5C2 = 10 ways to select 2 remaining points.

However, three of these 10 ways will give you a triangle that does not overlap (i.e. the points are successive).

If you have 6 points in a circle, and choose your first one, you can then choose two to the left OR two to the right OR one on either side, ie. there are three ways.

Hence, the number of non-overlapping triangles is
$image=https://latex.codecogs.com/png.image?\dpi{110}%20\frac{3}{10}\times%209240=2772&hash=4e0d8579c04dc320219ab5f45a7e5032$

ii) Similarly, the number of overlapping triangles is
$image=https://latex.codecogs.com/png.image?\dpi{110}%20\frac{7}{10}\times%209240=6468&hash=5105d2be780de0fc9ba2ae4ec06803dd$

Question 4 - part (d):

Subsequently, 2 + 6 + 24 + 24 = 56, matching the answer provided in your textbook.

I hope this helps!

Gtsh

New Member
Where did the 5 come from in '5C2 = 10 ways'.

5uckerberg

Active Member
I need help with these questions, I'm so confused

1. The diagram shows a 6 × 4 grid. The aim is to walk from the point A in the top left-hand corner to the point B in the bottom right-hand corner by walking along the black lines either downwards or to the right. A single move is defined as walking along one side of a single small square, thus it takes you ten moves to get from A to B.
Find how many different routes are possible:
I) without restriction,
ii) if you must pass through C

2. Twelve points are arranged in order around a circle.
b) In how many pairs of such triangles are the vertices of the two triangles distinct?
c) In how many such pairs will the triangles: i not overlap, ii overlap?

3. Find how many arrangements of the letters of the word TRANSITION are possible if
a) an N occupies the first but not the last position
b) the letter N is not at either end,
c) the vowels are together.

4. Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.
a)How many such numbers are there?
b) How many of them are odd?
c) How many of them are divisible by 5?
d) How many of them are divisible by 3?
3. The approach will go like this Since both Ns are the same then what we are working with is that suppose we put one N at the start and leave it alone and then there are 8 positions to put the next N so there we will have 8 times the something that we will explore later on. By doing so we have eight letters left which are T, R, A, S, I, T, I, O. Notice that the Ts and Is are repeated once. Then we will see that in the end, the solution for Q3 part a is
$\bg_white \frac{8\times{8!}}{2!2!}$

part b is simply stating now that the 2 Ns cannot be at the start at the end then we have a different situation then we will have 8 times 7 because first and foremost we are treating the Ns as unique to one another. From my point of view, the position of the Ns does not matter as we have the case of thing 1 and 2. The last remaining letters are worked out as the previous part. Thus this will give us $\bg_white \frac{8\times{7}\times{8!}}{2!2!}$.

part c is asking us to compile the fact that we have 10 choose 4 so thus we will have $\bg_white \frac{10!}{6!4!\times{2}}$ and then that result will have to be multiplied by 6 because this arrangement can be done in 6 different positions, then afterwards we will just do $\bg_white \frac{6!}{2!}$. In turn, this should give us $\bg_white \frac{10!6!}{6!4!2!2!}$

4a)Since you are given the fact that the number has to be less than 4000 so thus, we will first have for the 4 digit numbers we will have 2 times 24 giving us 12. Note the three, two and one digits will come later.

Speaking of this now enter the three-digit numbers we have $\bg_white 5\times{4}\times{3}=60$ and then for the two-digit numbers we have a total of 20 possibilities and then 5 possibilities for 1 digit numbers so in total it becomes $\bg_white 48 + 60 + 20 + 5 = 133$

b) Now similar to part a what will happen is that make 3 or 1 be the first number for the four-digit cases. Thus starting with $\bg_white 2\times{...}$. Then you are told it is an odd number so then from the given numbers 1, 3, 5, 8, 9. You see that if we used 1 or 3 we will have 3 options left so now it is $\bg_white 2 \times {3}\times{...}=6\times{...}$. After this is done you are left with 3 numbers in 2 spaces which give us $\bg_white 3\times{2}$ as there are 3 choices for the first space and then 2 choices for the second space. Thus, from the 4 digit numbers 36 cases.

With the 3 digit numbers, we will have to start with the condition that the number is odd. In that case, there are 4 possibilities for the last digit and then for the first 2 digits it can be anything so thus you are getting $\bg_white 4\times{12}=48$

For the two-digit numbers the start is similar to the 3 digit numbers which means 4 cases for the last digit and then 4 cases for the first digit thus giving us $\bg_white 4\times{4}=16$.

The 1 digit cases it is just 4. Adding all the digit cases together we have $\bg_white 36 + 48 + 16 + 4 = 104$. 104 cases.

Part c simply starts with the fact that a 3 or a 1 must be in the thousands category and then a 5 at the end and afterwards decide which 2 numbers from the 3 go in. This gives us 12 options. Four-digit condition.

Three-digit condition anchor the 5 there. and then you have 4 numbers to fill in 2 spaces. and we find that it is 12 cases.

Two-digit conditions do the same for the three-digit condition and then only 4 options left thus 4 cases.

One digit condition only one option.

Adding them together $\bg_white 12 + 12 + 4 + 1 = 29$

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