SOLUTION: HOW MANY PERMUTATIONS ARE THERE OF THE FOLLOWING WORD. M n1 MATHEMATICS 11 LETTERS n=11 A n2 T n3 H n4 E n5 M n6 ( 11 )

Question 480557: HOW MANY PERMUTATIONS ARE THERE OF THE FOLLOWING WORD.
M n1 MATHEMATICS 11 LETTERS n=11
A n2
T n3
H n4
E n5
M n6 ( 11 ) 11!
C n7 _____
T n8 ( 2,2,2,1,1,1,1,1 )= 2!2!2!1!1!1!1!1! now the question is what
I n9 do I do after this?
C n10 PLEASE SOMEONE EXPLAIN TO
S n11 ME STEP BY STEP:: THANKS
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mathematics has 11 letters.
there are 2 m's
there are 2 a's
that are 2 t's
the number of permutations should be:
11! / (2!*2!*2!) = 4989600
to see how this works, use much smaller numbers.
suppose your letters were "mat"
the number of permutations would be 3! = 6
those permutations would be:
mat
mta
amt
atm
tam
tma
now suppose your letters were "mma"
the number of permutations would be 3!/2! because 2 of the letters were the same.
those permutations would be:
mma
mam
amm
now suppose your letters were "mmat"
the number of permutations, if our equation is correct, would be 4!/2! = 12
those permutations would be:
mmat
mmta
mamt
matm
ammt
amtm
mtam
mtma
tmam
tmma
atmm
tamm
that's a total of 12.
now, suppose your letters are "mmtt"
the number of permutations would be 4!/(2!*2!) = 24/4 = 6
those permutations would be:
mmtt
ttmm
mtmt
tmtm
tmmt
mttm
that's a total of 6.
the formula works as can be seen with the smaller numbers.
applying it to the larger numbers gets you
number of permutations in "mathematics" is equal to 11!/(2!*2!*2!) = 4989600.

11! / (2!*2!*2!) is equal to 11*10*9*8*7*6*5*4*3*2*1 / 2*1*2*1*2*1
since the denominator winds up being 8, we cancel the 8 from the numerator with it to get:
11! / (2!*2!*2!) = 11*10*9*7*6*5*4*3*2*1
this is equivalent to:
11! / (2!2!*2!) = 110 * 96 * 30 * 12 * 2 which is equal to 6930 * 720 which is equal to 4989600

formula looks good and that should be your answer.

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