Question 3 & 4 Review Exercise 6

Solutions of Question 3 & 4 of Review Exercise 6 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

${ }^{56} P_{r+6}:{ }^{54} P_{r+3}=30800: 1$. Find $r$.

Given that: \begin{align} { }^{56} P_{r+6}:{ }^{54} P_r+3&=30800: 1 \\ \Rightarrow \dfrac{\dfrac{56 !}{[56-(r+6)] !}}{\dfrac{54 !}{[54-(r+3)] !}}&=\dfrac{30800}{1} \\ \Rightarrow \dfrac{56 !}{(50-r) !} \times \dfrac{(51-r) !}{54 !}&=30800 \\ \Rightarrow \dfrac{56.55 .54 !}{(50-r) !} \times \dfrac{(51-r)(50-r) !}{54 !}& =30800 \\ \Rightarrow \dfrac{3080}{1} \times \dfrac{51-r}{1}&=30800\\ \Rightarrow 51-r&=\dfrac{30800}{3080} \\ \Rightarrow r&=51-10=41\end{align}

In how many distinct ways can $x^4 y^3 z^5$ can be expressed without exponents?

We can write the word $x^4 y^3 z^5$ as: $$x^4 y^3 z^5=x . x . x . x . y . y . y . z . z . z . z . z $$ The total number of letter in this word are twelve,

so $n=12$ out of which four are $x$ so, $m_1=4$,

three are $y$ so $m_2=3$, and five are $z$, so $m_3=5$.

Thus total number of ways that $x^4 y^3 z^5$ can be arrange are \begin{align} & \left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)=\left(\begin{array}{c} 12 \\ 4,3,5 \end{array}\right)&=\dfrac{12 !}{4 ! 3 ! 5 !} \\ & =\dfrac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{4 ! 3 ! 5 !}=27,720 \end{align}

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