- WE WANT TO EVALUATE THE GIVEN EXPRESSIONS
INVOLVING FACTORIALS,
SO LET'S START WITH A REVIEW OF FACTORIALS.
FOR N FACTORIAL, N MUST BE A POSITIVE INTEGER
WHERE N! IS THE PRODUCT OF POSITIVE INTEGERS
LESS THAN OR = N.
SO NOTICE HOW FOR N! WE START WITH A FACTOR OF N.
EACH FACTOR DECREASES BY 1
UNTIL WE END WITH THE FACTORS OF 2 AND 1.
SO FOR EXAMPLE, 6! = 6 x 5 x 4 x 3 x 2 x 1.
IT'S ALSO TRUE THAT N! CAN BE WRITTEN AS N(N - 1)!,
SO FOR EXAMPLE, 8! = 8 x 7! BECAUSE IF WE EXPANDED 7!
WE WOULD HAVE 7 x 6 x 5 x 4 x 3 x 2 x 1
AND BY DEFINITION, 0! = 1.
FACTORIALS OCCUR IN TOPICS SUCH AS COMBINATORICS
AS WELL AS THE BINOMIAL THEOREM,
BUT FOR THIS VIDEO WE'RE GOING TO FOCUS
ON JUST EVALUATING EXPRESSIONS INVOLVING FACTORIALS.
SO 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1.
OF COURSE, WE COULD FIND THIS PRODUCT BY HAND,
BUT LET'S GO AHEAD AND USE A CALCULATOR.
MANY SCIENTIFIC CALCULATORS DO HAVE A FACTORIAL KEY,
SO WE'LL GO AHEAD AND SHOW THOSE TWO WAYS.
FIRST, DO IT THE LONG WAY.
OF COURSE, WE COULD LEAVE OFF THE FACTOR OF 1, BUT 7! = 5,040
OR USING THE FACTORIAL KEY,
ON THIS CALCULATOR WE CAN PRESS 7, MATH,
ARROW OVER TO PRB FOR PROBABILITY,
OPTION 4 IS FACTORIAL, AND ENTER.
OF COURSE, THE RESULT IS THE SAME.
LET'S GO AHEAD AND TAKE A LOOK AT OUR SECOND EXAMPLE.
THERE ARE A COUPLE WAYS OF SIMPLIFYING THIS,
SO WE'LL GO AHEAD AND SHOW BOTH.
FIRST, LET'S GO AHEAD AND EXPAND 8! AND 6!.
SO WE'D HAVE 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1/6!
WHICH WOULD BE 6 x 5 x 4 x 3 x 2 x 1.
NOTICE WE CAN SIMPLIFY THIS FRACTION.
6/6 SIMPLIFIES TO 1, 5/5 SIMPLIFIES TO 1, AND SO ON.
ALL THESE SIMPLIFY TO 1/1,
THEREFORE, THE RESULT IS JUST 8 x 7 WHICH = 56.
ANOTHER WAY TO GET THE SAME RESULT
THAT WOULD REQUIRE LESS WORK
WOULD BE TO RECOGNIZE THAT IF OUR DENOMINATOR IS 6!
AND THE NUMERATOR IS 8!
WE CAN WRITE 8! AS 8 x 7 x INSTEAD OF EXPANDING THIS,
WE COULD WRITE x 6!.
IN THIS FORM, NOTICE THAT 6!/6! WOULD SIMPLIFY TO 1
GIVING US THE SAME RESULT OF 8 x 7 OR 56.
AND THEN FOR THE LAST EXAMPLE,
AGAIN, WE CAN APPROACH THIS TWO WAYS.
FIRST, APPROACH IT THE LONG WAY BY EXPANDING EACH FACTORIAL.
NOTICE THE DENOMINATOR HAS TWO FACTORIALS HERE,
SO WE'LL FIRST HAVE 7!,
AND THEN WE HAVE 3! WHICH WOULD BE 3 x 2 x 1.
AND NOW WE'LL GO AHEAD AND SIMPLIFY--7/7, 6/6, AND SO ON.
NOW NOTICE WE HAVE 10 x 9 x 8 IN THE NUMERATOR,
AND WE HAVE 3 x 2 x 1 IN THE DENOMINATOR.
WELL, NOTICE THAT 9 AND 3 SHARE A COMMON FACTOR OF 3,
SO THERE'S 1 3 IN 3, AND 3 3'S IN 9.
AND ALSO THE 8 AND THE 2 SIMPLIFY
BECAUSE THEY SHARE A COMMON FACTOR OF 2.
THERE'S 1 2 IN 2 AND 4 2'S IN 8.
NOTICE HOW THIS SIMPLIFIES NICELY TO 10 x 3 x 4
WHICH WOULD BE 120.
LET'S SHOW THIS ANOTHER WAY.
AGAIN, NOTICING THAT WE HAVE 10! IN THE NUMERATOR
AND 7! IN THE DENOMINATOR,
WE COULD WRITE 10! AS 10 x 9 x 8 x 7!,
AND THEN FOR THE DENOMINATOR, WE WOULD HAVE 7!.
AND THEN WE'LL GO AHEAD AND EXPAND 3!
WHICH WOULD BE 3 x 2 x 1.
NOTICE HOW THIS REQUIRES LESS WORK.
7!/7! SIMPLIFIES OUT,
AND THEN AGAIN WE HAVE A COMMON FACTOR BETWEEN THE 3 AND THE 9.
THERE'S 1 3 IN 3 AND 3 3'S IN 9,
AND THERE'S A COMMON FACTOR OF 2 BETWEEN 8 AND 2.
THERE'S 1 2 IN 2 AND 4 2'S IN 8, SO THE RESULT IS THE SAME.
WE HAVE 10 x 3 x 4 WHICH IS 120.
LET'S GO AHEAD AND VERIFY THIS ONE ON THE CALCULATOR.
WE HAVE 10!.
PRESS MATH, LEFT ARROW ONCE TO PROBABILITY, OPTION 4,
DIVIDED BY, IN PARENTHESES WE HAVE 7! x 3!.
WE NEED THESE PARENTHESES
TO MAKE SURE BOTH OF THESE FACTORIALS
ARE IN THE DENOMINATOR,
SO WE HAVE 7, MATH, LEFT ARROW, OPTION 4, AND THEN 3!, 3, MATH,
LEFT ARROW, OPTION 4, CLOSE PARENTHESIS, AND ENTER.
SO WE HAVE SIMPLIFIED THIS CORRECTLY.
I THINK WE'LL GO AHEAD AND STOP HERE FOR THIS VIDEO.
I HOPE YOU FOUND THIS HELPFUL.