Question

find the number of PERMUTATIONS that can be formed from the letters of the word POPULAR and how many have two P's separated?

Answer 1

if you had 7 different letters it would be \(7!\) by the counting principle. since you cannot tell the "P"s apart, it is \(\frac{7!}{2}\)

Answer 2

well that gives the correct answer! thanks!!!

Answer 3

good. now for the second part, i think we have to find the number of permutations where they ARE together and subtract that from the previous answer.
does that seem right?

Answer 4

yessss! plus, stationary salesman sell 1 pencil case, 1 eraser, 2 pens, 3 pencils,, howmany different sequences can these be sold?

Answer 5

we can consider the 2 "P' s as one letter to compute that number.
then it is like having 6 letters, and the number of ways to permute 6 letters is \(6!\)

Answer 6

thank you, im really havin problem on this!!

Answer 7

then your answer to part B should be
\[\frac{7!}{2}-6!\] if i am not mistaken

Answer 8

i get 1,800
http://www.wolframalpha.com/input/?i=7!%2F2-6!
that is for part B of the first question
now i will look at the second one

Answer 9

i am not sure i understand this question
stationary salesman sell 1 pencil case, 1 eraser, 2 pens, 3 pencils, how many different sequences can these be sold?

Answer 10

if it means he can sell them in any order, but you cannot tell the pens apart, and cannot tell the pencils apart, then it is
\[\frac{6!}{2!3!}\]

Answer 11

does that look right?

Answer 12

i am not sure if you can check your answers on line, but i interpret the question correctly it should be 60

Answer 13

the answer is 420 in my scheme answer :/

Answer 14

then i did not interpret the question correctly
i wonder what it is actually asking

Answer 15

oh i guess it would help if i could count! you have 7 items, not 6, so it is
\[\frac{7!}{2!3!}=420\]

Answer 16

i miscounted the number of items, sorry

Answer 17

oh yes i got it , it is
7!\[\frac{ 7! }{ 2!3! }\]

Answer 18

more importantly, is it clear how this works?

Answer 19

yess.,, thank you :)

Answer 20

yw

Answer 21

anyway for the first question, for the POPULAR, how many ways if 2 Ps are separated? because the answer on the top of the tread was for the "no of permutations can be formed from POPULAR"

Answer 22

there are two parts to this question, right?

Answer 23

1) number of permutations with no restriction we got as \(\frac{7!}{2!}=2520\)

Answer 24

actually there is 4 parts of it,
1)no of permutations can be formed from POPULAR words
2) howmany ways begin and end with p
3)how many have 2P's separated
4)how many of them have vowels together

Answer 25

and how is the working for no 3??

Answer 26

for 3) it would be the number of ways with no restrictions, minus the number of ways with the two P together. that gives the total number of ways with the two separated.
i answered that but will be happy to write it again

Answer 27

we already computed the number of ways with no restrictions, it is
\[\frac{7!}{2!}\]
if the two P are together you can consider them as on letter "(PP)" and then it is like having 6 letters
the number of ways to permute 6 letters is \(6!\)
your answer is therefore
\[\frac{7!}{2!}-6!\]

Answer 28

yes!i overlooked, im sorry and thank you again,:)

Answer 29

you good with the last one too?

Answer 30

yup, i managed to answer that. hehe

Answer 31

oh i thought that might be the trickiest
don't you find some of these annoying? i do

Answer 32

yeah, so annoying. i always that this topic is for geniuses! how you managed to get the working correctly? besides the fact that you are also a genius? is there any tricks or way to study them?sorry for the long post!