Question 1173440: How many 3-letter words can be formed from the letters in these words.
PEDRO=
Can you help me with this? thank you so much Found 2 solutions by Theo, MathTherapy:Answer by Theo(13342) (Show Source): You can put this solution on YOUR website! the formula you will need are:
c(n,x) = n! / (x! * (n-x)!)
p(n,x) = n! / (n-x)!
the first is the combination formula.
it tells you the number of sets of x you can make from a set of n where order is not important.
the second is the permutation formula.
it tells you the number of sets of x you can make from a set of n where order is important.
if order is not important, then each set has to have elements that are different regrdless of the position that they are in the set.
if order is important, then each set can have the same elements as another set but in a different order.
an example would be to find a set of 2 elements from a set of 3 elements.
the number of combinations would be c(3,2) = 3! / (2! * 1!) = (3*2*1) / (2*1*1) = 3
the number of permutations would be p(3,2) = 3! / (1!) = (3*2*1) / (1) = 6
assume the lettes are ABC.
the 3 possible combinations would be:
AB
AC
BC
notice that no set has the same elements as any other set.
the 3 possible permutations are:
AB
BA
AC
CA
BC
CB
notice that more than 1 set can have the same elements in it as another set.
they cannot, however, be in the same order.
example
BC and CB have the same elements, but they are not in the same order.
if order was not important, then BC and CB would be considered members of the same set, i.e. the combination formula produces set BC or the set CB but would not produces the sets BC and CB.
back to your problem *************************
with PEDRO, you would first want to find the number of sets where order is not important.
that would be C(5,3) = 5! / (3! * 2!) = (5*4*3*2*1) / (3*2*1*2*1) = (5*4) / (2*1) = 20/2 = 10
those 10 sets would be:
PED ***** 1
PER ***** 2
PEO ***** 3
PDR ***** 4
PDO ***** 5
PRO ***** 6
EDR ***** 7
EDO ***** 8
ERO ***** 9
DRO ***** 10
none of the above sets have all 3 elements that are the same as elements in any other set, regardless of the order of the elements within each set.
the number of permutations would be P(5,3) = 5! / (2!) = (5*4*3*2*1) / (2*1) = (5*4*3) = 60.
that would be the 10 possible sets where at least one element in each set is different from the other sets times 6 possible arrangements within each set.
taking your 6 possible combinations and finding the 6 possible arrangements within each of them would result in:
PED ***** 1
PDE
DEP
DPE
EPD
EDP
PER ***** 2
PRE
ERP
EPR
RPE
REP
PEO ***** 3
POE
EOP
EPO
OPE
OEP
PDR ***** 4
PRD
DRP
DPR
RPD
RDP
PDO ***** 5
POD
DOP
DPO
OPD
ODP
PRO ***** 6
POR
ROP
RPO
OPR
ORP
EDR ***** 7
ERD
DRE
DET
RED
RDE
EDO ***** 8
EOD]
DOE
DEO
OED
ODE
ERO ***** 9
EOR
ROE
REO
OER
ORE
DRO ***** 10
DOR
ROD
RDO
ODR
ORD
your answer should be 60, since they want to know how many possible 3 letter words can you make from the letters PEDRO.
it is assumed that order is important.
PED is a different word from DEP.
you would use the permutation formula of P(5,3).
these formulas assume that the same letter can't be repeated more than once.
in other words, PPP would not be considered since the letter P can only be used once.