How do you simplify $(16!)/(9!*7!)$ ?

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Answer 1 · 2016-01-31T14:57:11

$16! =1*2*3*4*5*6*7.......*16$

$9! =1*2*3*4*5*6*7...*9$

$7! =1*2*3*4*5*6*7$

$(16!)/(9!*7!)=$

$(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16)/((1*2*3*4*5*6*7*8*9)(1*2*3*4*5*6*7))$

The number that equal other cancel out!

$rarr=(10*11*12*13*14*15*16)/(1*2*3*4*5*6*7)$

$7$ and $14$ cancel out , $6$ and $12$ cancel out , $5$ and $10$ cancel out!

$rarr=(2*11*2*13*2*15*16)/(1*2*3*4*1*1*1)$

Now $4$ and $14$ cancel out, $2$ and $2$ cancel out, $3$ and $15$ cancel out!

$rarr=(2*11*1*13*2*5*4)/(1*1*1*1*1*1*1)$

$rarr=2*11*1*13*2*5*4$

$rarr=22*26*20$

$rarr=440*26$

$rarr=11440$

Answer 2 · 2016-02-07T21:21:53

11440

$(16!)/(9!*7!)=( 16*15*14*13*12*11*10*cancel(9!))/(cancel(9!)*7!)$

$( 16*cancel(15)*14*13*12*11*10)/(7*6*cancel(5)*4*cancel(3)*2*1)$

$( 16*cancel(14)*13*12*11*10)/(cancel(7)*6*4*cancel(2)*1)$

$(16*13*6*2*11*(2*5))/(6*2*2)$

$16*13*11*5$

11440

How do you simplify (16!)/(9!*7!)(16!)/(9!*7!)?