perms and combs q (1 Viewer)

[anonymoushehe]

What am I doing wrong the answers for a is 480 and b is 360.

 

[alewe]

For (a): you know that M is included, but the number of combinations of other letters you can choose is 6C3 to get your 4 letters to rearrange. Then once you get your 4 letters, they can be arranged in 4! ways, so the solution should be 6C3*4!=480

 

[alewe]

(b) should be 360 because you know you have to have the O and M, but you still need to select 2 more letters from the remaining 6 letters in 6C2 ways. Then once you have the 4 letters they can be arranged in 4! ways, so the solution should be 6C2*4!=360

 

[liamkk112]

View attachment 43862What am I doing wrong the answers for a is 480 and b is 360.

a) if M is included and C is not, then we are simply arranging DOMESTI and including M. firstly we know that M is included, so we need not worry about counting that. then, we have 6 other letters to choose 3 from, so 6C3 ways to choose those letters. then we can arrange the letters in 4! ways, so overall there is 4! 6C3 = 480 ways to do this.
b) similarly, we are simply arranging DOMESTIC and including O and M. we know that O and M are included so we don't worry about that, and there are 6C2 ways to choose the remaining 2 letters for, and then 4! ways to arrange all the letters. this gives 4! 6C2 = 360 ways to do this

 

[anonymoushehe]

a) if M is included and C is not, then we are simply arranging DOMESTI and including M. firstly we know that M is included, so we need not worry about counting that. then, we have 6 other letters to choose 3 from, so 6C3 ways to choose those letters. then we can arrange the letters in 4! ways, so overall there is 4! 6C3 = 480 ways to do this.
b) similarly, we are simply arranging DOMESTIC and including O and M. we know that O and M are included so we don't worry about that, and there are 6C2 ways to choose the remaining 2 letters for, and then 4! ways to arrange all the letters. this gives 4! 6C2 = 360 ways to do this

ohh so for my working out in a was I just assuming the case that the first or last letter was M?

 

[liamkk112]

ohh so for my working out in a was I just assuming the case that the first or last letter was M?

i think so, u just arranged the remaining three letters but didn't shuffle the M around

 

[ISAM77]

a) There are 4 spaces: _ _ _ _
You must put M in one of the four spaces = 4 possible ways.
The remaining three spaces may be filled by: D,O,E,S,T,I = 6*5*4 = 120 possible ways.
4*120 = 480 possible ways.

Think of it as: there are 120 possible ways to choose 3 different letters from D,O,E,S,T,I. And We multiply it by 4 possible ways we could insert M into that mix of 3 letters to make 4 letters in total.

b) There are 4 spaces: _ _ _ _
O and M must be in two of the four spaces. You can arrange O and M in 4 * 3 = 12 possible ways.
The remaining two spaces are filled by D,E,S,T,I,C. You can arrange them in 6 * 5 = 30 possible ways
12*30 = 360 possible ways.

Think of it as: there are 12 possible ways to choose spots for O and M. And We multiply this by the 30 possible ways we can insert the other letters into the mix to make 4 letters in total.

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That's my way of thinking for this question. Hope it helps!

If this doesn't help; you could start writing the possible arrangements of letters and think of how to make a "tree of possible outcomes" with them to picture it better. That's how I would continue explaining my logic if this isn't enough.