Which Is Larger?

Probability Level 3

Let

  • ρ \rho = number of ways you can arrange the 6 letters of R R , 5 letters of G G , 4 letters of J J , 3 letters of M M and 2 letters of K K to create a single word.

  • μ \mu = number of ways you can arrange the 4 letters of R R , 3 letters of G G , 4 letters of J J , 3 letters of M M , 2 letters of K K , 1 letter of A A and 1 letter of B B to create a single word.

Relate ρ \rho and μ \mu .

For more problems like this, try answering this set .

μ < ρ \mu < \rho ρ < μ \rho < \mu μ = ρ \mu = \rho

Christian Daang by Christian Daang , 20, Philippines

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1 solution

Christian Daang
Apr 3, 2017

ρ = ( 6 + 5 + 4 + 3 + 2 ) ! 6 ! ⋅ 5 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! = 20 ! 6 ! ⋅ 5 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! \rho = \cfrac{(6 + 5 + 4 + 3 + 2)!}{6! \cdot 5! \cdot 4! \cdot 3! \cdot 2!} = \cfrac{20!}{6! \cdot 5! \cdot 4! \cdot 3! \cdot 2!}

while

μ = ( 4 + 3 + 4 + 3 + 2 + 1 + 1 ) ! 4 ! ⋅ 3 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! = 18 ! 4 ! ⋅ 3 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! \mu = \cfrac{(4 + 3 + 4 + 3 + 2 + 1 + 1)!}{4! \cdot 3! \cdot 4! \cdot 3! \cdot 2! \cdot 1! \cdot 1! } = \cfrac{18!}{4! \cdot 3! \cdot 4! \cdot 3! \cdot 2! \cdot 1! \cdot 1! }

Now,

20 ! 6 ! ⋅ 5 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! _?_ 18 ! 4 ! ⋅ 3 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! ( 20 ⋅ 19 ) _?_ ( 6 ! ⋅ 5 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! ) ( 4 ! ⋅ 3 ! ⋅ 4 ! ⋅ 3 ! ⋅ 2 ! ⋅ 1 ! ⋅ 1 ! ) 380 < 5 ⋅ 6 ⋅ 5 ⋅ 4 = 600 ∴ ρ < μ \begin{aligned} \cfrac{20!}{6! \cdot 5! \cdot 4! \cdot 3! \cdot 2!} \ & \text{\_?\_} \ \cfrac{18!}{4! \cdot 3! \cdot 4! \cdot 3! \cdot 2! \cdot 1! \cdot 1! } \\ (20 \cdot 19) \ & \text{\_?\_} \ \cfrac{(6! \cdot 5! \cancel{\cdot 4! \cdot 3! \cdot 2!} )}{(4! \cdot 3! \cdot \cancel{4! \cdot 3! \cdot 2!} \cdot 1! \cdot 1!)} \\ 380 \ & \text{<} \ 5 \cdot 6 \cdot 5 \cdot 4 = 600 \\ \therefore & \boxed{\rho < \mu} \end{aligned}

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