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): are next to each other
(b): are not next to each other
(c): 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

(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 \$

(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 \$

(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 ${}_2 \Pi_{5} = 2^5 = 32$ .

(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

. 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 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 $\frac{8!}{4! 2! 1! 1!}$ ．