Chapter 13.1, Problem 51E

a.

To determine

To calculate: The number of different ways to select 4 plants among corn, wheat and bean plants.

Answer to Problem 51E

The number of different ways to select 4 plants among corn, wheat and bean plantsare $330$ .

Explanation of Solution

Given information:

There are 6 corn, 3 wheat and 2 bean plants.

Formula used:

If there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Calculation:

Consider the provided information that there are 6 corn, 3 wheat and 2 bean plants.

Total number of plants are,

  $6+3+2=11$

In order to select 4 plants at random, use combination as selection does not matter.

Recall that if there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Required number of ways are $C\left(11,4\right)$ .

Here n is 11 and r is 4for $C\left(11,4\right)$ .

  $\begin{array}{c}C\left(11,4\right)=\frac{11!}{4!\left(11-4\right)!}\\ =\frac{11!}{4!\cdot 7!}\\ =\frac{11\cdot 10\cdot 9\cdot 8\cdot 7!}{4!\cdot 7!}\\ =330\end{array}$

Thus, the number of different ways to select 4 plants among corn, wheat and bean plants are $330$ .

b.

To determine

To calculate: The number of different ways to select 4 plants among corn, wheat and bean plants if exactly 2 corn plants must be included.

Answer to Problem 51E

The number of different ways to select 4 plants among corn, wheat and bean plants if exactly 2 corn plants must be includedare $150$ .

Explanation of Solution

Given information:

There are 6 corn, 3 wheat and 2 bean plants.

Formula used:

If there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Calculation:

Consider the provided information that there are 6 corn, 3 wheat and 2 bean plants.

Total number of plants are,

  $6+3+2=11$

Out of 4 2 must be corn plants.

Number to ways to select 2 plants from 6 corn plants are $C\left(6,2\right)$ as selection does not matter.

In order to select 2 remaining plants 5 plants are there 3 of wheat and 2 of bean, use combination as selection does not matter.

Number of ways are $C\left(5,2\right)$ .

Recall that if there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Required number of ways are $C\left(6,2\right)\cdot C\left(5,2\right)$ .

Here n is 6 and r is 2 for $C\left(6,2\right)$ and n is 5 and r is 2 for $C\left(5,2\right)$ .

  $\begin{array}{c}C\left(6,2\right)\cdot C\left(5,2\right)=\frac{6!}{2!\left(6-2\right)!}\cdot \frac{5!}{2!\left(5-2\right)!}\\ =\frac{6\cdot 5\cdot 4!}{2!\cdot 4!}\cdot \frac{5\cdot 4\cdot 3!}{2!\cdot 3!}\\ =\frac{3\cdot 5}{1}\cdot \frac{5\cdot 2}{1}\\ =150\end{array}$

Thus, the number of different ways to select 4 plants among corn, wheat and bean plants if exactly 2 corn plants must be included are $150$ .