Trial Paper Q (2 Viewers)

[krypticlemonjuice]

For this question, and in the second photo, the solution, can someone please explain how they expanded it to get that??? I'm confused

 

[synthesisFR]

i did this paper yesterday howd u go

 

[Alpharex83]

If you're talking about the highlighted part:
For LHS, you use the binomial theorem formula, and on RHS, you expand nCk into factorial form and multiply it by 1/n^k. The reason why it ends at (n-k+1) is because the remaining terms cancel out with (n-k)! in the denominator

 

[SB257426]

in line 4 of my working: i expanded n! in numerator and (n-k)! in denominator and u can see that i cancelled out the terms... thats how the answer comes out to be

 

[synthesisFR]

View attachment 40489

in line 4 of my working: i expanded n! in numerator and (n-k)! in denominator and u can see that i cancelled out the terms... thats how the answer comes out to be

dont laugh at me how do u do part ii

 

[synthesisFR]

is it just like u take them being opposite as one count, so then its 7 ways so its just 6!
thats so wierd

 

[SB257426]

lock the position of the host and the hostess and so there are 2! ways of placing the host and hostess (they can swap)

now look at the remaining 4 people. they can be arranged normally in 4! ways

if i am correct then its 4! times 2! = 48 ways

otherwise i suck at combinatorics (i actually hate this topic ngl)

 

[Alpharex83]

is it just like u take them being opposite as one count, so then its 7 ways so its just 6!
thats so wierd

Yes I'm pretty sure it's 6!. You lock in the host and hostess, and then arrange the guests around them. The guests can be arranged in 6! ways. The reason why we don't multiply 6! by 2! (due to the misconception that the host and hostess swapping places will give new cases) is because of this:

Let the other guests be A, B, C, D, E, and F, host is H and hostess is S

Consider the case HABCSDEF. Swapping H and S would give SABCHDEF. However, this case is already accounted for in the 6! in the case HDEFSABC (remember, it's a round table).

Thus, 6! should be the answer.

 

[SB257426]

lock the position of the host and the hostess and so there are 2! ways of placing the host and hostess (they can swap)

now look at the remaining 4 people. they can be arranged normally in 4! ways

if i am correct then its 4! times 2! = 48 ways

otherwise i suck at combinatorics (i actually hate this topic ngl)

oh my mistake i thought for a second the host and hostess were included in the 6 people !

then it would be 6! times 2 !

 

[Luukas.2]

It is 6!, not 6! x 2.

Seat the host in any seat in 1 way (all seats are equivalent under rotation).

The seven seats remaining are now all different, but there is only one of them in which the hostess can be seated.

The remaining 6 people can be seated in 6! ways.

So, # ways = 1 x 1 x 6! = 6!

 

[=)(=]

View attachment 40488View attachment 40487


For this question, and in the second photo, the solution, can someone please explain how they expanded it to get that??? I'm confused

wait which paper is this?

 

[synthesisFR]

wait which paper is this?

Nsg 2023

 

[=)(=]

ohh right ic cheers

 

[SB257426]

It is 6!, not 6! x 2.

Seat the host in any seat in 1 way (all seats are equivalent under rotation).

The seven seats remaining are now all different, but there is only one of them in which the hostess can be seated.

The remaining 6 people can be seated in 6! ways.

So, # ways = 1 x 1 x 6! = 6!

my bad.. you are correct