question (1 Viewer)

[yellowhighlighterr]

part d help anyone?

 

[yellowhighlighterr]

ik that the last digit must be 0 or 5, leaving us with the ss now being 8 values.
so p(div by 5) = 8*7*6*1 (if its 0) and the same for 5 (=336+
they both give 336, so wouldn't you just multiply 336 by 2 together to get the answer????

 

[tywebb]

It is cambridge y11 14B Q15

 

[yellowhighlighterr]

It is cambridge y11 14B Q15

View attachment 39054
View attachment 39055

but if a number ends in 0 it is also divisible by 5 right??? so the answer should be 672???

 

[tywebb]

It won't end in 0. The digits are 1,2,3,..,9 which does not include 0.

 

[yellowhighlighterr]

heres a weird one
part d
any help pls?

 

[yellowhighlighterr]

In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

 

[synthesisFR]

View attachment 39056

heres a weird one
part d
any help pls?

just take (NU) as one letter and dont do 2! inside bc thats the only arrangement
then just normal arrangement

 

[synthesisFR]

wait actually my trials r soon and i swear i havent done perms and combs since like prelims

 

[yanujw]

heres a weird one
part d
any help pls?

There are two approaches. First you could think that either N is to the left of U or it isn't, and there is no condition that makes one possibility more likely so there are half as many ways as if there were no restrictions. Thus 6!/2 = 360 ways

Or, visualise just N and U already being there next to each other, and that the other letters are to be placed around them so that N always remains before U. There are 3 positions for the first letter placed, 4 for the next, then 5 and then 6 positions for the final letter placed. Apply the multiplication principle for a total of 360 ways.

just take (NU) as one letter and dont do 2! inside bc thats the only arrangement
then just normal arrangement

This approach is not quite correct, this places N and U next to each other rather than the premise of the question.

 

[yellowhighlighterr]

There are two approaches. First you could think that either N is to the left of U or it isn't, and there is no condition that makes one possibility more likely so there are half as many ways as if there were no restrictions. Thus 6!/2 = 360 ways

Or, visualise just N and U already being there next to each other, and that the other letters are to be placed around them so that N always remains before U. There are 3 positions for the first letter placed, 4 for the next, then 5 and then 6 positions for the final letter placed. Apply the multiplication principle for a total of 360 ways.


This approach is not quite correct, this places N and U next to each other rather than the premise of the question.

could u elaborate further pls?

 

[synthesisFR]

There are two approaches. First you could think that either N is to the left of U or it isn't, and there is no condition that makes one possibility more likely so there are half as many ways as if there were no restrictions. Thus 6!/2 = 360 ways

Or, visualise just N and U already being there next to each other, and that the other letters are to be placed around them so that N always remains before U. There are 3 positions for the first letter placed, 4 for the next, then 5 and then 6 positions for the final letter placed. Apply the multiplication principle for a total of 360 ways.


This approach is not quite correct, this places N and U next to each other rather than the premise of the question.

OHHHHH that makes sense
yep ignore what i said i was thinking like N and U have to be like right next to each other

this proves my point that i need to relearn this topic before trials

 

[yellowhighlighterr]

i am starting to hate perms and combs and i just started-

 

[yellowhighlighterr]

11b



12c

 

[tywebb]

View attachment 39056

heres a weird one
part d
any help pls?

This is yr11 cambridge 14B Q18

 

[tywebb]

In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

This is yr11 cambridge 14B Q19

 

[tywebb]

View attachment 39057

i am starting to hate perms and combs and i just started-

These are yr 11 cambridge 14B Q22-24

 

[tywebb]

View attachment 39061

11b

View attachment 39062

12c

These are yr 11 cambridge 14A Q10-12

 

[yellowhighlighterr]

This has been a massive help
ik that i can count on bos for maths help

 

[yellowhighlighterr]

part a please?