Permutation, Combination

Exercise 1
1. Find the following number of 4-digit integer when arranging 7 numbers from 0 to 6.
(a)
all numbers are different.
(b)
multiple of 5.
(c)
number can be repeated.
2. When arranging 6 cards from the cards numbered from 1 to 10, find the number of outcomes in the following cases.
(a)
all arrangements
(b)
including 1 and 2
(c)
including 1 or 2
3. When tossing a coin 5 times, find the number of outcomes in the following cases.
(a)
The head shows up none,1,2,3,4,5 times
(b)
all cases
4. Find the number of outcomes when arranging $a,b,c,d,e,f$ in the following sequence.
(a)
$a,b$ are next to each other
(b)
$a,b$ are not next to each other
(c)
$a,b$ are place to the each end

5. Show that the number of outcomes when arranging $a,a,a,a,b,b,c,d$ in a sequence is given by ${8 \choose 4,2,1,1}$.

Answer

1

(a) Consider the number of outcomes by putting the integers in $\square \ \square \ \square \ \square$. First of all, it has to be 4-digit integer. Thus we can not use 0 in the thousands. Then we have a choice of one of integers. Thus, there are 6 outcomes. For hundreds through ones, we can use any of the integers. But we can not use the same integer twice. Then there are 6 outcomes in the hundreds, 5 outcomes in the tens, 4 outcomes in the ones. Thus, total of

$\displaystyle 6\cdot6\cdot5\cdot4 = 720 $

(b) Note that multiples of 5 always have a ones digit of 0 or 5.Next, note that the case where the ones digit is 0 and the case where it is 5 are considered separately

When the ones place is 0.

Thousands, hundreds, and tens can use numbers from 1 to 6 once. The number of permutations to take out 3 out of 6 is ${}_6 P_{3}$.Therefore, 120

When the ones place is 5.

We can not use 0 in the thousands. Thus, there are 5 outcomes in the thousands. There are ${}_5P_{2}$ outcomes for the hundreds and tens and $5\cdot {}_5P_{2} = 5\cdot 5 \cdot 4 = 100$ outcomes in the ones. Thus,

$\displaystyle 120 + 100 = 220 $

(c) Here we can use the same number. Now the are 6 outcomes in the thousands and 7 outcomes in the hunreds, tens, and ones.,Therefore,

$\displaystyle 6\cdot {}_7 \Pi_{3} = 6 \cdot 7 \cdot 7 \cdot 7 = 2058 \ $

2

(a) The number of outcomes for taking 6 cards from 10 different cards is ${}_{10} C_{6}$. Now

$\displaystyle {}_{10} C_{6} = \frac{10!}{4! 6!} = \frac{10 \cdot 9 \cdot 8 \cdot 7 }{4 \cdot 3\cdot 2\cdot 1} = 210 \ $

(b) Cards 1 and 2 have to be in the arrangement. So, we consider the number of outcomes of selecting 4 cards from 8 cards. Thus we have ${}_8 C_{4} = 70$

(c) Let $C_{1}$ be the case where 1 is included and $C_{2}$ be the case where 2 is included. Then there are ${}_9 C_{5}$ outcomes in $C_1$ and ${}_9 C_{5}$ outcomes in $C_2$. Thus, the outcomes for which the card 1 or 2 is included is

$\displaystyle {}_9 C_{5} + {}_9 C_{5} - 70 = 252 - 70 = 182 \ $

3

(a) $A_{i} = [number of heads is i]$.Then the heads can be shown $i$ times whithin 5 trials. Thus, the number of outcomes is ${}_5 C_{i}$

The number of outcomes of $A_{0}$ is ${}_5 C_{0} = 1$$A_{1}$ is ${}_5 C_{1} = 5$$A_{2}$ is ${}_5 C_{2} = 10$$A_{3}$ is ${}_5 C_{3} = 10$$A_{4}$ is ${}_5 C_{4} = 5$$A_{5}$ is ${}_5 C_{5} = 1$

(b) In all possible cases, the number of outcomes is ${}_2 \Pi_{5} = 2^5 = 32$.

4

(a) Since $a,b$ are nect to each other,let $A = a,b$. Then we have $A,c,d,e,f$.Thus, the nuber of outcomes is

$\displaystyle {}_5 P_{5} = 5! = 120 $

Also, let $B = b,a$. Then the number of outcomes is

$\displaystyle {}_5 P_{5} = 5! = 120 $

Therefore, total of $240$.

(b) Let $E = $[$a,b$ are not next to each other]. Then $\bar{E} = $[$a,b$ are next to each other]. Thus the number of outcomes of $E = $

$\displaystyle 6! - 240 = 720 - 240 = 480 \ $

(c) $a,b$ are at the ends. Then there are two cases. case 1. $a,c,d,e,f,b$. Then the number of arrangements of $c,d,e,f$ is ${}_4 P_{4}$ case 2. $b,c,d,e,f,a$. Then the number of arrangements of $c,d,e,f$ is ${}_4 P_{4}$. Thus,

$\displaystyle 2 \cdot {}_4 P_{4} = 2 \cdot 4! = 48 \ $

5 ${8 \choose 4,2,1,1} = \frac{8!}{4! 2! 1! 1!}$.If you number all and arrange them, then the number of outcomes is ${}_8 P_{8} = 8!$.But there are 4 $a$'s. 2 $b$'s. They can not be distinguished. So, the number of outcomes is $\frac{8!}{4! 2! 1! 1!}$