Given the word "pelleted", how many arrangements can be made if the 3 e's are together?
You would treat the e's as 1 letter and do 6! / 2! = 360 ways right?
My 2nd question is, from a group of 10 children, 6 boys and 4 girls, how many ways can a group of 4 people be selected if:
a) no restriction: C(10,4)=210
b) 4 boys: C(6,4) = 15
c) 4 girls: C(4,4) = 1
d) 1 girl: C(6,3) + C(4,1) = 24
e) 2 girls: C(6,2) + C(6,2) = 21
f) 3 girls: C(6,1) + C(4,3) = 10
I was thinking that these were right at first, but after I finished them I tried to add them all up thinking that they should = a) which is 210. They don't. Did I do them wrong? Are they not suppose to add up to the one with no restriction?
You would treat the e's as 1 letter and do 6! / 2! = 360 ways right?
My 2nd question is, from a group of 10 children, 6 boys and 4 girls, how many ways can a group of 4 people be selected if:
a) no restriction: C(10,4)=210
b) 4 boys: C(6,4) = 15
c) 4 girls: C(4,4) = 1
d) 1 girl: C(6,3) + C(4,1) = 24
e) 2 girls: C(6,2) + C(6,2) = 21
f) 3 girls: C(6,1) + C(4,3) = 10
I was thinking that these were right at first, but after I finished them I tried to add them all up thinking that they should = a) which is 210. They don't. Did I do them wrong? Are they not suppose to add up to the one with no restriction?