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Number Theory Level pending

m ! = ( 64 32 ) × ( 32 16 ) 2 × ( 16 8 ) 4 × ( 8 4 ) 8 × ( 4 2 ) 16 × ( 2 1 ) 32 m! = \dbinom{64}{32}\times\dbinom{32}{16}^2 \times \binom{16}{8}^4 \times\dbinom{8}{4}^8\times\dbinom{4}{2}^{16}\times\dbinom{2}{1}^{32}

Find the value of m m satisfying the equation above.

Clarification: The notation ( x y ) \binom{x}{y} indicates "x choose y" or the binomial coefficient indexed by x and y.


The answer is 64.

Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
Apr 23, 2016

( 2 n n ) = ( 2 n ) ! / ( n ! ∗ n ! ) \binom{2n}{n} = (2n)!/(n!*n!) So, everything cancels out except for 64!. So, m = 64 m = \boxed{64}

i.e.

m ! = ( 64 32 ) ∗ ( 32 16 ) 2 ∗ ( 16 8 ) 4 ∗ ( 8 4 ) 8 ∗ ( 4 2 ) 16 ∗ ( 2 1 ) 32 = 64 ! 32 ! 2 ∗ 32 ! 2 16 ! 4 ∗ 16 ! 4 8 ! 8 ∗ 8 ! 8 4 ! 16 ∗ 4 ! 16 2 ! 32 ∗ 2 ! 32 1 ! 64 = 64 ! m! = \binom{64}{32}*\binom{32}{16}^2*\binom{16}{8}^4 *\binom{8}{4}^8*\binom{4}{2}^{16}*\binom{2}{1}^{32} = \frac{64!}{32!^2} * \frac{32!^2}{16!^4} * \frac{16!^4}{8!^8} * \frac{8!^8}{4!^{16}} * \frac{4!^{16}}{2!^{32}} * \frac{2!^{32}}{1!^{64}} = 64!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Plz detail it

MRITYUYNJAY KUMAR - 5 years, 1 month ago

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OK, I added a more detailed explanation above... :)

Geoff Pilling - 5 years, 1 month ago

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