Question

A shipment of $10$ microwaves contains $3$ defective units. In how many ways can a vending company purchase $4$ of these units and receive (a) all good units (b) two good units and (c) at least two good units?

Answer 1

To solve this question, we need to find out the number of ways of receiving the good units and the defective units in each case. Since the good and the defective units are received simultaneously, so by the multiplication theorem, the total number of ways will be obtained by multiplying the number of ways of receiving good and defective units. The calculation for the number of ways will be done with the help of the combination formula, which is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Complete step by step answer:
According to the question, the total number of units of microwaves contained in the shipment is equal to $10$, out of which $3$ are the defective ones. This means that the number of good units in the shipment is equal to $7$ units.
(a)
According to the question, the vending company is receiving all good units in this case. As written above, the total number of good units in the shipment is equal to $7$. Since the company is purchasing $4$ units, this means that it is purchasing $4$ good units out of the $7$ good units present in the shipment.
Now, the number of ways in which this can happen is equal to the total number of combinations of $4$ good units out of the $7$ good units. So the required number of ways in this case is
$\begin{aligned} & \Rightarrow {{n}_{1}}={}^{7}{{C}_{4}} \\\ & \Rightarrow {{n}_{1}}=\dfrac{7!}{4!\left( 7-4 \right)!} \\\ & \Rightarrow {{n}_{1}}=\dfrac{7!}{4!\times 3!} \\\ \end{aligned}$
Writing $7!=7\times 6\times 5\times 4!$ in the above equation, we get

& \Rightarrow {{n}_{1}}=\dfrac{7\times 6\times 5\times 4!}{4!\times 3!} \\\ & \Rightarrow {{n}_{1}}=\dfrac{7\times 6\times 5}{3!} \\\ \end{aligned}$$ We know that $$3!=3\times 2\times 1=6$$. Substituting this above, we get $$\begin{aligned} & \Rightarrow {{n}_{1}}=\dfrac{7\times 6\times 5}{6} \\\ & \Rightarrow {{n}_{1}}=7\times 5 \\\ & \therefore {{n}_{1}}=35 \\\ \end{aligned}$$ Therefore the number of ways in this case is equal to $$35$$. (b) In this case, the vending company is receiving two good units out of the total four units purchased. This means that it is receiving two defective units. The total number of good units in the shipment is equal to $7$. So the number of ways for receiving $2$ good units out of $7$ good units is equal to ${}^{7}{{C}_{2}}$. Also, the total number of defective units in the shipment is equal to $3$. So the number of ways of receiving $2$ defective units out of $3$ defective units is equal to $^{3}{{C}_{2}}$. Since both of these events are occurring simultaneously, so by the multiplication theorem the total number of ways in this case is given by $$\begin{aligned} & \Rightarrow {{n}_{2}}={}^{7}{{C}_{2}}\times {}^{3}{{C}_{2}} \\\ & \Rightarrow {{n}_{2}}=\dfrac{7!}{2!\left( 7-2 \right)!}\times \dfrac{3!}{2!\left( 3-2 \right)!} \\\ & \Rightarrow {{n}_{2}}=\dfrac{7\times 6\times 5!}{2!\times 5!}\times \dfrac{3\times 2!}{2!\times 1!} \\\ & \Rightarrow {{n}_{2}}=\dfrac{7\times 6}{2!}\times \dfrac{3}{1!} \\\ \end{aligned}$$ We know that $$2!=2\times 1=2$$ and $$1!=1$$. Putting these above, we get $$\begin{aligned} & \Rightarrow {{n}_{2}}=\dfrac{7\times 6}{2}\times \dfrac{3}{1} \\\ & \Rightarrow {{n}_{2}}=21\times 3 \\\ & \therefore {{n}_{2}}=63 \\\ \end{aligned}$$ Therefore the number of ways in this case is equal to $$63$$. (c) In this case, the vending company is receiving at least two good units. Mathematically, it means that the company receives two or more than two good units, out of the total $4$ units purchased. This in turn means that the number of ways in this case is equal to the sum of the number of ways of receiving $2$, $3$, and $4$ good units out of the total $7$ good units. In these respective cases, the number of defective units received is equal to $2$, $1$, and $0$ respectively out of the total $3$ defective units. So the number of ways is given by $$\begin{aligned} & \Rightarrow {{n}_{3}}={}^{7}{{C}_{2}}\times {}^{3}{{C}_{2}}+{}^{7}{{C}_{3}}\times {}^{3}{{C}_{1}}+{}^{7}{{C}_{4}}\times {}^{3}{{C}_{0}} \\\ & \Rightarrow {{n}_{3}}=\dfrac{7!}{2!\left( 7-2 \right)!}\times \dfrac{3!}{2!\left( 3-2 \right)!}+\dfrac{7!}{3!\left( 7-3 \right)!}\times \dfrac{3!}{1!\left( 3-1 \right)!}+\dfrac{7!}{4!\left( 7-4 \right)!}\times \dfrac{3!}{0!\left( 3-0 \right)!} \\\ \end{aligned}$$ On solving we get $$\therefore {{n}_{3}}=203$$ **Therefore the number of ways in this case is equal to $$203$$.** **Note:** Do not forget to multiply the number of ways of receiving the defective units with the number of ways of receiving the given number of good units in each of the above cases. In the first case, all of the purchased units were good. So we ignored the calculation for the defective units in that case.