- WELCOME TO A VIDEO ON PERMUTATIONS.
THE GOALS OF THIS VIDEO ARE TO EVALUATE FACTORIALS,
USE PERMUTATIONS TO SOLVE PROBLEMS AND ALSO TO DETERMINE
THE NUMBER OF PERMUTATIONS WITH INDISTINGUISHABLE ITEMS.
A PERMUTATION IS AN ARRANGEMENT OR ORDERING OF ITEMS.
SO THE MAIN THING TO REMEMBER FOR A PERMUTATION
IS THAT THE ORDER MATTERS.
SO IF WE WANT TO KNOW HOW MANY DIFFERENT WAYS 3 STUDENTS
CAN LINE UP TO PURCHASE A NEW TEXTBOOK READER,
THIS WOULD BE A PERMUTATION SINCE THE ORDER MATTERS.
SO IF WE LABEL THE 3 STUDENTS "A," B AND C,
LET'S SEE IF WE CAN LIST THE POSSIBLE ARRANGEMENTS.
WELL, WE COULD HAVE "A," B, C.
AND THEN IF STUDENT B AND C CHANGED ORDER,
WE WOULD HAVE "A," C, B.
NOW, WE'LL LET STUDENT B BE FIRST,
SO WE'D HAVE B, "A," C OR B, C, "A."
AND THEN LASTLY IF STUDENT C IS FIRST,
WE'D HAVE C, "A," B OR C, B, "A."
SO WE'D SAY THE NUMBER OF PERMUTATIONS WOULD BE 6.
NOW, WE CAN MODEL THE SITUATION IN ANOTHER WAY.
YOU CAN THINK OF THE 3 POSITIONS IN LINE.
AND IF WE TAKE A LOOK AT THE FIRST POSITION IN LINE,
THERE WOULD BE 3 WAYS IT COULD
BE FILLED WITH STUDENT "A," B OR C.
NOW ONCE THE FIRST POSITION IN LINE IS FILLED,
THERE WOULD ONLY BE 2 POSSIBILITIES
FOR THE SECOND POSITION.
AND THEN LASTLY, ONCE THE FIRST 2 POSITIONS ARE FILLED,
THERE'S ONLY 1 OPTION FOR THAT LAST POSITION.
SO USING THE COUNTING PRINCIPLE IDEA,
IF WE MULTIPLY 3 x 2 x 1 THAT WOULD ALSO GIVE US
6 POSSIBLE WAYS THAT 3 STUDENTS CAN LINE UP.
NOW, THIS BRINGS US TO THE IDEA OF FACTORIAL
N FACTORIAL GIVES THE NUMBER OF PERMUTATIONS FOR N ITEMS,
AND THIS IS THE NOTATION FOR N FACTORIAL.
N FACTORIAL = N x N - 1 x N - 2, ALL THE WAY DOWN TO x 2 x 1.
NOW, N DOES HAVE TO BE A WHOLE NUMBER AND BY DEFINITION,
0 FACTORIAL = 1.
SO IF WE TAKE A LOOK AT 3 FACTORIAL,
THIS MEANS WE START WITH A FACTOR OF 3
AND THEN WE'RE GOING TO MULTIPLY BY FACTORS THAT DECREASE BY 1
UNTIL WE END UP WITH A FACTOR OF 1.
SO WE'D HAVE 3 x 2 x 1 WHICH = 6.
SO 6 FACTORIAL WOULD BE 6 x 5 x 4 x 3 x 2 x 1,
AND THIS WOULD BE 720.
LET'S GO AHEAD AND SHOW WHERE FACTORIAL BUTTON
IS ON CALCULATOR.
IF I WANTED TO DETERMINE LET'S SAY 12 FACTORIAL,
I COULD JUST TYPE IN 12, PRESS MATH, ARROW OVER TO PROBABILITY,
AND IT'S OPTION 4.
AND THERE WE GO.
AND REMEMBER THAT N FACTORIAL
DOES REPRESENT THE NUMBER OF WAYS
WE CAN PERMUTE OR ORDER N DISTINCT ITEMS.
DISTINCT, MEANING WE CAN TELL THEM APART.
LET'S TAKE A LOOK AT PERMUTATION NOTATION.
PERMUTATIONS OF N ITEMS TAKEN R AT A TIME IS WRITTEN
USING THIS NOTATION AND AGAIN REPRESENTS
A NUMBER OF PERMUTATIONS OF N ITEMS R AT A TIME.
SO IF YOU TAKE A LOOK AT THIS NOTATION HERE,
THIS IS THE PERMUTATION OF 7 ITEMS TAKEN 3 AT A TIME.
THIS IS A PERMUTATION OF 15 ITEMS TAKEN 5 AT A TIME.
AND THIS IS EQUAL TO N FACTORIAL DIVIDED BY N - R FACTORIAL,
AND SOMETIMES THIS IS ALSO WRITTEN N PERMUTE R
AS WELL DEPENDING ON WHAT TEXTBOOK YOU READ.
SO IF WE WANT TO EVALUATE THE PERMUTATIONS OF 7 ITEMS
TAKEN 3 AT A TIME, IT WOULD BE 7 FACTORIAL
DIVIDED BY N - R FACTORIAL.
WELL, 7 - 3 FACTORIAL, THAT = 7 FACTORIAL/4 FACTORIAL.
LET'S GO AHEAD AND EXPAND THIS.
THIS WOULD BE 7 x 6 ALL THE WAY DOWN TO x 1 ALL/4 x 3 x 2 x 1.
AND YOU'LL SEE THIS DOES SIMPLIFY 4/4 IS 1, 3/3 IS 1,
AND SO ON, SO THIS = 7 x 6 x 5 WHICH = 210.
I'D LIKE TO PAUSE FOR A MOMENT
AND TAKE A LOOK AT THIS PROBLEM AGAIN.
IF WE HAVE 7 ITEMS AND WANT TO PERMUTE 3 OF THEM AT A TIME,
WE COULD MODEL THIS WITH 3 POSITIONS.
AND WE HAVE 7 CHOICES FOR THE FIRST POSITION,
6 FOR THE SECOND, AND 5 FOR THE FOURTH
WHICH AS EXPECTED GIVES US THE SAME RESULT OF 210.
OKAY, FOR THE SECOND QUESTION, WE HAVE 15 PERMUTE 5,
THAT = 15 FACTORIAL/15 - 5 FACTORIAL
WHICH = 15 FACTORIAL/10 FACTORIAL.
LET'S GO AHEAD AND START EXPANDING THIS.
NOW NOTICE FROM HERE ON OUT,
THAT WOULD BE 10 x 9 x 8 AND SO ON,
SO THAT WOULD BE 10 FACTORIAL.
AND THE REASON I'M STOPPING THERE
IS BECAUSE THE DENOMINATOR IS ALSO 10 FACTORIAL,
SO WE'LL NOTICE THAT
10 FACTORIAL/10 FACTORIAL SIMPLIFIES OUT.
FOR THE SAME REASON UP HERE,
WE COULD'VE LEFT THIS AS 4 FACTORIAL AND 4 FACTORIAL.
LET'S GO AHEAD AND USE OUR CALCULATOR
TO DETERMINE THIS PRODUCT, THAT = 360,000,360.
OKAY, LET'S TAKE A LOOK AT A COUPLE PROBLEMS NOW.
THE CLASS HAS 28 STUDENTS.
HOW MANY DIFFERENT ARRANGEMENTS CAN 5 STUDENTS
GIVE A PRESENTATION TO THE CLASS?
WELL, THE ORDER DOES MATTER, SO THIS IS A PERMUTATION.
SO WE HAVE THE PERMUTATION OF 28 ITEMS TAKEN 5 AT A TIME,
SO THAT'S 28 FACTORIAL/28 - 5 FACTORIAL,
THAT = 28 FACTORIAL/23 FACTORIAL.
AND I ALSO WANT TO SHOW YOU HOW YOU CAN EVALUATE A PERMUTATION
IN THIS FORM ON THE GRAPHING CALCULATOR.
AND THIS IS WHY SOMETIMES YOU MIGHT PREFER TO USE THE NOTATION
THAT I SHOWED ON THE PREVIOUS SCREEN AS 28 PERMUTE 5 ITEMS.
THIS IS MORE OF HOW WE TYPE IT INTO THE CALCULATOR.
SO IN ORDER TO EVALUATE THIS ON THE CALCULATOR,
WE WOULD HAVE TO TYPE IN 28, PRESS MATH OVER TO PROBABILITY,
AND YOU'LL SEE THE PERMUTATION OPTION AS OPTION 2.
AND THEN WE HAVE TO TYPE IN THE VALUE OF R
WHICH WAS 5, 11,793,600.
LET'S GO AHEAD AND EVALUATE ON THIS FORM, TOO,
TO SHOW THAT IT WOULD RESULT IN THE SAME VALUE.
SO WE HAVE 28 FACTORIAL, REMEMBER THAT WAS UNDER MATH,
PROBABILITY, OPTION 4, DIVIDED BY 23 FACTORIAL, OPTION 4,
AND YOU CAN SEE THE NUMBER OF PERMUTATIONS
IS OBVIOUSLY THE SAME.
HOW MANY WAYS CAN THE LETTERS OF THE WORD PHOENIX BE ARRANGED?
WELL, THE FIRST THING WE HAVE TO NOTICE IS THAT EVERY LETTER
IN THIS WORD IS DIFFERENT,
AND THAT'S IMPORTANT AND WE'LL SEE IN JUST A MOMENT.
SO WE HAVE 7 DIFFERENT OR 7 DISTINCT LETTERS.
AND EACH TIME WE ARRANGE THEM, WE WANT TO USE ALL THE LETTERS.
SO THIS WOULD BE A PERMUTATION OF 7 ITEMS TAKEN 7 AT A TIME.
AGAIN, IT CAN BE WRITTEN IN THIS FORM AS WELL,
AND IT = N FACTORIAL, 7 FACTORIAL/7 - 7 FACTORIAL,
THAT'S 0 FACTORIAL.
AND BY DEFINITION, 0 FACTORIAL = 1,
SO THIS IS 7 x 6 x 5 x 4 x 3 x 2 x 1.
SO LET'S GO BACK TO THE CALCULATOR.
WE'RE GOING TO HAVE 7 FACTORIAL, OR WE COULD TYPE IN 7, MATH,
PROBABILITY, OPTION 2, 7 PERMUTE, 7 ITEMS.
AND OF COURSE, THE RESULT IS THE SAME, 5,040.
LET'S TAKE A LOOK AT ONE MORE IDEA.
SOMETIMES WE HAVE A PERMUTATION WHERE THE ITEMS
ARE INDISTINGUISHABLE OR WE CAN'T TELL THEM APART
WHICH MEANS THAT WHEN WE SWITCH THE ORDER OF THE ITEMS,
WE WOULDN'T BE ABLE TO TELL
BECAUSE YOU CAN'T TELL THE ITEMS APART.
AND HERE'S HOW WE'LL ADDRESS THIS TYPE OF PERMUTATION.
THE NUMBER OF DIFFERENT PERMUTATIONS OF N OBJECTS
WHERE THERE ARE N SUB 1 INDISTINGUISHABLE ITEMS,
N SUB 2 INDISTINGUISHABLE ITEMS
AND SO ON IS GIVEN BY THIS FORMULA.
SO WE'LL TAKE N FACTORIAL
AND THEN DIVIDE BY N SUB 1 FACTORIAL, N SUB 2 FACTORIAL
AND SO ON, WHERE N SUB 1 AND N SUB 2
ARE THE NUMBER OF INDISTINGUISHABLE ITEMS
OF EACH TYPE.
FOR EXAMPLE, HOW MANY WAYS CAN THE LETTERS OF THE WORDS
MATHEMATICS BE REARRANGED?
SO WE'LL FIRST COUNT THE NUMBER OF LETTERS IN TOTAL,
SO WE HAVE 11 DIFFERENT LETTERS.
AND NOW, WE HAVE TO COUNT,
AND NOW WE HAVE TO DETERMINE THE INDISTINGUISHABLE ITEMS.
WELL, THESE 2 Ms ARE INDISTINGUISHABLE,
SO WE MUST DIVIDE BY 2 FACTORIAL SINCE THERE'S 2 Ms.
LET'S GO AHEAD AND CROSS THOSE OFF, SO WE KEEP TRACK OF THIS.
NOTICE THERE'S ALSO 2 As,
SO WE HAVE TO DIVIDE BY ANOTHER 2 FACTORIAL, AND ANYTHING ELSE?
LOOKS LIKE THERE'S ALSO 2 Ts,
SO WE HAVE TO DIVIDE BY 2 FACTORIAL.
NOW IF THERE WERE 3 OF ANYTHING, WE WOULD DIVIDE BY 3 FACTORIAL.
SO WHAT WE'RE DOING HERE IS WE'RE DIVIDING OUT
THE PERMUTATIONS THAT WE WOULD NOT BE ABLE TO TELL
THEY WERE DIFFERENT BECAUSE OF THE LETTERS
THAT LOOK EXACTLY THE SAME.
LET'S GO AHEAD AND FIND THIS VALUE.
SO WE HAVE 11 FACTORIAL, AND WE'RE GOING TO DIVIDE THIS BY,
SINCE 2 FACTORIAL IS 2 x 1, THE DENOMINATOR WOULD BE 8.
LET'S JUST DIVIDE BY 8 AND THERE WE HAVE IT,
4,989,600 DIFFERENT PERMUTATIONS.
LET'S SEE IF WE CAN DO ONE MORE.
HOW MANY WAYS CAN YOU ORDER
3 BLUE MARBLES, 4 RED MARBLES AND 5 GREEN MARBLES?
AGAIN THE ORDER MATTERS, SO IT'S A PERMUTATION.
BUT THE MARBLES OF THE SAME COLOR LOOK IDENTICAL,
SO THEY'RE INDISTINGUISHABLE.
SO WE HAVE A TOTAL OF 3 + 4 + 5 MARBLES
THAT WOULD BE 12 MARBLES, SO WE HAVE 12 FACTORIAL.
AND THEN SINCE THERE ARE 3 INDISTINGUISHABLE BLUE MARBLES,
WE DIVIDE BY 3 FACTORIAL, 4 INDISTINGUISHABLE RED MARBLES,
SO WE DIVIDE BY 4 FACTORIAL,
AND THE SAME FOR THE 5 GREEN MARBLES.
SO THIS QUOTIENT WILL TELL US THE NUMBER OF ARRANGEMENTS
THAT WE CAN MAKE WITH THESE MARBLES
GIVEN THE MARBLES WITH THE SAME COLOR LOOK EXACTLY THE SAME.
SO WE HAVE 12 FACTORIAL IN THE NUMERATOR,
AND WE'LL DIVIDE THIS
BY 3 FACTORIAL x 4 FACTORIAL x 5 FACTORIAL.
SO WE HAVE 27,720 DIFFERENT WAYS WE CAN ORDER THESE 12 MARBLES.
OKAY, I HOPE YOU FOUND THE VIDEO HELPFUL.
HAVE A GOOD DAY.