why do we add 1? (1 Viewer)

[kkk579]

^^

 

[kkk579]

for this q how on earth do u have a fraction for the number of ways

 

[kkk579]

Bruh im so confused for this q why do we use C instead of P cause it says to "arrange",doesn't that mean theres order? Also for another q which states to form words out any APPLES, they also use C instead of P which makes no sense since words also have order this topic genuine bullshit or maybe ngo is js shit at teaching it idk

Q: in how many ways can i select 2math books from 6 math books and 2 eco books from 4 eco books and arrange them on a shelf in any order

 

[t234]

for this q how on earth do u have a fraction for the number of ways

part d ii is a probability so it can be a fraction :>

 

[kkk579]

part d ii is a probability so it can be a fraction :>

Oh yeah thanks for that didnt read properly

 

[liamkk112]

^^View attachment 42935

pick 2 of the 8 points to draw a line between. Then we subtract 3C2 to get rid of collinear duplicates. but we now are missing the 1 case where we have the line between the 3 collinear points, since we just subtracted that off by getting rid of 3C2, as that's still a possibility so we need to add the 1 back on. tricky

 

[kkk579]

pick 2 of the 8 points to draw a line between. Then we subtract 3C2 to get rid of collinear duplicates. but we now are missing the 1 case where we have the line between the 3 collinear points, since we just subtracted that off by getting rid of 3C2, as that's still a possibility so we need to add the 1 back on. tricky

wdym by collinear duplicates?

 

[liamkk112]

wdym by collinear duplicates?

draw out those 3 collinear points, call them A, B, C. the line between A and B, the line between B and C, the line between A and C are all the same since they lie on the same line. So we subtract these 3C2 duplicate lines (we choose 2 out of the 3 collinear points). but now we still need to include the line through A,B,C, so we add 1 to recover this

 

[kkk579]

draw out those 3 collinear points, call them A, B, C. the line between A and B, the line between B and C, the line between A and C are all the same since they lie on the same line. So we subtract these 3C2 duplicate lines (we choose 2 out of the 3 collinear points). but now we still need to include the line through A,B,C, so we add 1 to recover this

Hmm lowkey im still kinda confused

 

[liamkk112]

Hmm lowkey im still kinda confused

consider the easier case when we have 3 collinear points (lets just take the points (1,1) and (2,2) and (3,3) for example, they all lie on the line y =x), and one point that isn't (we will take (4,2) which doesn't lie on y =x), giving 4points in total. we have 4C2 ways to create a straight line from any 4 points: we choose 2 out of the 4 points to draw a line between them, and repetition isn't allowed because the line from say (1,1) to (4,2) is the exact same as the line from (4,2) to (1,1) for example, so that's why we are using choose here to avoid these repeats. This gives 4C2 = 6 "naive" possible lines ignoring our collinear restrction: the line from (1,1) to (2,2), the line from (1,1) to (3,3), the line from (1,1) to (4,2), the line from (2,2) to (3,3), the line from (2,2) to (4,2), and finally the line from (3,3) to (4,2).

however, we can immediately see here that we have overcounted by 3 because the line from (1,1) to (2,2) is the exact same as the line from (1,1) to (3,3) which is the exact same as the line from (2,2) to (3,3). the 3 here comes from choosing 2 out of the 3 collinear points to draw a line between, so we can say that there are 3C2 ways in general, again using choose here for the same reason as above.

so if we take out these possibilities where we overcounted , we end up with 4C2-3C2 = 3 possible lines, namely the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). but now we can see that the line that the 3 collinear points lied on is missing from this list, aka the line y = x. this is still a valid possiblity, because it's still a line that connects 2 of our 4 possible points, it's just that it happened to be counted 3 times. so we need to add back on the line y = x to our possible lines.

so we now see that there are 4C2-3C2 + 1 = 4 possible lines in total, namely y =x, the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). you can then scale this up to include as many points as you want, same logic applies

 

[kkk579]

consider the easier case when we have 3 collinear points (lets just take the points (1,1) and (2,2) and (3,3) for example, they all lie on the line y =x), and one point that isn't (we will take (4,2) which doesn't lie on y =x), giving 4points in total. we have 4C2 ways to create a straight line from any 4 points: we choose 2 out of the 4 points to draw a line between them, and repetition isn't allowed because the line from say (1,1) to (4,2) is the exact same as the line from (4,2) to (1,1) for example, so that's why we are using choose here to avoid these repeats. This gives 4C2 = 6 "naive" possible lines ignoring our collinear restrction: the line from (1,1) to (2,2), the line from (1,1) to (3,3), the line from (1,1) to (4,2), the line from (2,2) to (3,3), the line from (2,2) to (4,2), and finally the line from (3,3) to (4,2).

however, we can immediately see here that we have overcounted by 3 because the line from (1,1) to (2,2) is the exact same as the line from (1,1) to (3,3) which is the exact same as the line from (2,2) to (3,3). the 3 here comes from choosing 2 out of the 3 collinear points to draw a line between, so we can say that there are 3C2 ways in general, again using choose here for the same reason as above.

so if we take out these possibilities where we overcounted , we end up with 4C2-3C2 = 3 possible lines, namely the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). but now we can see that the line that the 3 collinear points lied on is missing from this list, aka the line y = x. this is still a valid possiblity, because it's still a line that connects 2 of our 4 possible points, it's just that it happened to be counted 3 times. so we need to add back on the line y = x to our possible lines.

so we now see that there are 4C2-3C2 + 1 = 4 possible lines in total, namely y =x, the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). you can then scale this up to include as many points as you want, same logic applies

OHHHH this makes so much more sense now tysm for the help!!

 

[kkk579]

consider the easier case when we have 3 collinear points (lets just take the points (1,1) and (2,2) and (3,3) for example, they all lie on the line y =x), and one point that isn't (we will take (4,2) which doesn't lie on y =x), giving 4points in total. we have 4C2 ways to create a straight line from any 4 points: we choose 2 out of the 4 points to draw a line between them, and repetition isn't allowed because the line from say (1,1) to (4,2) is the exact same as the line from (4,2) to (1,1) for example, so that's why we are using choose here to avoid these repeats. This gives 4C2 = 6 "naive" possible lines ignoring our collinear restrction: the line from (1,1) to (2,2), the line from (1,1) to (3,3), the line from (1,1) to (4,2), the line from (2,2) to (3,3), the line from (2,2) to (4,2), and finally the line from (3,3) to (4,2).

however, we can immediately see here that we have overcounted by 3 because the line from (1,1) to (2,2) is the exact same as the line from (1,1) to (3,3) which is the exact same as the line from (2,2) to (3,3). the 3 here comes from choosing 2 out of the 3 collinear points to draw a line between, so we can say that there are 3C2 ways in general, again using choose here for the same reason as above.

so if we take out these possibilities where we overcounted , we end up with 4C2-3C2 = 3 possible lines, namely the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). but now we can see that the line that the 3 collinear points lied on is missing from this list, aka the line y = x. this is still a valid possiblity, because it's still a line that connects 2 of our 4 possible points, it's just that it happened to be counted 3 times. so we need to add back on the line y = x to our possible lines.

so we now see that there are 4C2-3C2 + 1 = 4 possible lines in total, namely y =x, the line from (1,1) to (4,2), the line from (2,2) to (4,2), and the line from (3,3) to (4,2). you can then scale this up to include as many points as you want, same logic applies

For this q, is it like this?

 

[liamkk112]

For this q, is it like this? View attachment 42947View attachment 42947View attachment 42948

looks right to me

 

[kkk579]

looks right to me

Alright thanks