combinations ext math question (1 Viewer)

qwerla

New Member
Twelve people arrive at a restaurant. There is one table for six, one table for four and one table for two. In how many ways can they be assigned to a table?
thank you

notme123

Active Member
What's the answer? I haven't done perms and combs in a while but my guess is (12C6)5!*(6C4)3!*(2C2)1! = 9979200.

Last edited:

qwerla

New Member
What's the answer? I haven't done perms and combs in a while but my guess is (12C6)5!*(6C4)3!*(2C2)1! = 9979200.
the answer is 13860 ( i can't seem to wrap my head around it bc i've tried everything and i still don't get the answer. thank u for trying anyway! perm and comb is so hardd

imaiyuki

Member

this is in my y11 notes haha
in this case we would have
$\bg_white {12!\over {6!4!2!}}=13860$
as seen in the last formula

hope this helps!

Trebla

Do it sequentially:

- There are $\bg_white \binom{12}{6}$ ways to select 12 people to form the table of 6
- Once those 6 are selected there are 6 left so there are $\bg_white \binom{6}{4}$ ways to select 6 people to form the table of 4
- Once those 4 are selected there are 2 left so there is $\bg_white \binom{2}{2}$ way to form the table of 2

CM_Tutor

Moderator
Moderator
@qwerla, Trebla's answer is correct (obviously). @notme123's answer is for a different question... in how many ways can the people be seated around tables of 6, 4, and 2, whereas the question asked about assigning people to tables.

notme123

Active Member
@qwerla, Trebla's answer is correct (obviously). @notme123's answer is for a different question... in how many ways can the people be seated around tables of 6, 4, and 2, whereas the question asked about assigning people to tables.
Love it when I misread the question lol.

qwerla

New Member
View attachment 30620
this is in my y11 notes haha
in this case we would have
$\bg_white {12!\over {6!4!2!}}=13860$
as seen in the last formula

hope this helps!
hey!!! thank you SO much! this makes sm sense to me now, and thank you so much for the picture above. it really clarified it for me. hope u have a great rest of ur day

qwerla

New Member
Do it sequentially:

- There are $\bg_white \binom{12}{6}$ ways to select 12 people to form the table of 6
- Once those 6 are selected there are 6 left so there are $\bg_white \binom{6}{4}$ ways to select 6 people to form the table of 4
- Once those 4 are selected there are 2 left so there is $\bg_white \binom{2}{2}$ way to form the table of 2