Solution

Because the order matters, the number of different orders in this race is the number of 5-permutations of a set of 5 elements. Consequently, the answers is $P(5, 5) = 5! = 120$.

Solution

Because the order matters, the number of different orders of names is the number of 6-permutations of a set of 6 elements. Consequently, the answers is $P(6, 6) = 6! = 720$.

Solution

Because the order of 1s doesn't matter, the number of bit strings of length $10$ contain exactly $n$ 1s can be given by the number of $n$-combinations of a set of $10$ elements, where $n \mid 0 \leqslant n \leqslant 10$. That is

$C(10, n) = \frac{10!}{(10-n)!n!}$

a. The number of bit string of length 10 contain exactly $4$ 1s is $C(10, 4) = \dfrac{10!}{(10-4)!4!} = 210$
b. By sum rule, the number of bit string of lenght 10 contains at most four 1s is the sum of numbers of bit strings contain exactly $n$ 1s, where is this case $0 \leqslant n \leqslant 4$.

$C_b = C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3) + C(10, 4)$

c. By sum rule, the number of bit string of lenght 10 contains at least four 1s is the sum of numbers of bit strings contain exactly $n$ 1s, where in this case $4 \leqslant n \leqslant 10$.

$C_c = C(10, 4) + C(10, 5) + \cdots + C(10, 10)$

d. The answer can be given by the number of bit strings of length of 10 contain exactly $5$ 1s, that is

$C_d = C(10, 5) = \frac{10!}{5!5!} = 252$

Solution

Because the order of 1s doesn't matter, the number of bit strings of length $12$ contain exactly $n$ 1s can be given by the number of $n$-combinations of a set of $12$ elements, where $n \mid 0 \leqslant n \leqslant 12$. That is

$C(12, n) = \frac{12!}{(12-n)!n!}$

a. The number of bit string of length 12 contain exactly $3$ 1s is $C(12, 3) = \dfrac{12!}{(12-3)!3!} = 220$
b. By sum rule, the number of bit string of lenght 12 contains at most three 1s is the sum of numbers of bit strings contain exactly $n$ 1s, where is this case $0 \leqslant n \leqslant 3$.

$C_b = C(12, 0) + C(12, 1) + C(12, 2) + C(12, 3)$

c. By sum rule, the number of bit string of lenght 12 contains at least three 1s is the sum of numbers of bit strings contain exactly $n$ 1s, where in this case $3 \leqslant n \leqslant 12$.

$C_c = C(12, 3) + C(12, 4) + \cdots + C(12, 12)$

d. The answer can be given by the number of bit strings of length of 12 contain exactly $6$ 1s, that is

$C_d = C(12, 6) = \frac{12!}{6!6!} = 924$

Solution

Because the order of chosen integers doesn't matter, the number of ways to choose 2 positive integers less than 100 is the number of $2$-combinations of a set of 99 elements. Consequently, the answer is

$C(99, 2)= \frac{99!}{97!2!} = 4851$

Solution

Let the number of elements of such one subset be $n$, where $n \in [1, 3, 5, 7, 9]$.
1. Because the order doesn't matter, the number of subsets with $n$ elements is the number of $n$-combinations of a set with 10 element.
2. By sum rule, the number of subsets with an odd number of elements of a set with 10 elements is the sum of the numbers of subsets with $n$ elements. That is

$C = C(10, 1) + C(10, 3) + C(10, 5) + C(10, 7)+ C(10, 9)$

Solution

a. By product rule, the total number is $2^{10}$.
b. Because the order of heads doesn't matter, the number of possible outcomes contain exactly two heads is the number of 2-combinations of a set of 10 elements. The answer is $C(10, 2) = \dfrac{10!}{8!2!} = 45$.
c. By sum rule, the number of possible outcomes contain at most three tails is the sum of numbers of possible outcomes contain exactly $n$ tails, where in this case $n\in [0, 1, 2, 3]$. That is

$C_c = C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3)$

d. The answer can be given by the number of possible outcomes contain exactly $5$ heads, that is

$C_d = C(10, 5) = \frac{10!}{5!5!} = 252$

Solution

a. We can treat ED as one element, then the answer can be given by the number of 7-permutations of a set of 7 elements. That is $P(7, 7) = 7! = 5040$.
b. We can treat CDE as one element, then the answer can be given by the number of 6-permutations of a set of 6 elements. That is $P(6, 6) = 6! = 720$.
c. We can treat BA and FGH as two elements, then the answer can be given by the number of 5-permutations of a set of 5 elements. That is $P(5, 5) = 5! = 120$.

Solution

a. Because the order of chosen members doesn't matter, the number of choosing 4 members from the club is the number of 4-combinations of a set of 25 elements. The answer is $C(25, 4) = \frac{25!}{21!4!}=12650$.
b. Because the order of chosen members matters, the number of choosing 4 members from the club is the number of 4-permutions of a set of 25 elements. The answer is $P(25, 4) = \frac{25!}{21!}=303600$.

Solution

1. We need to choose 3 women from 15 women and 3 men from 10 men.
2. Because the order doesn't matter, the number $C_{w}$ of ways to choose 3 women from 15 women is the number of 3-combinations of a set of 15 elements. The number $C_{m}$ of ways to choose 3 men from 10 women is the number of 3-combinations of a set of 10 elements. It follows that

$\begin{aligned} C_{w} &= C(15, 3) = \frac{15!}{12!3!} = 455\\ C_{m} &= C(10, 3) = \frac{10!}{7!3!} = 120 \end{aligned}$

3. By product rule, the number of ways to form such committee is $C_{w} \cdot C_{m} = 455 \cdot 120 = 54600$

Solution

Let the numbers of women and men in the committee be $w$ and $m$ respectively, where $w+m=6$ and $w>m$. It follows that

$[w,m] \in \{[6, 0], [5, 1], [4, 2]\}$

1. The number $C_w$ of ways to choose $w$ women from 15 women is the number of $w$-combinations of a set of 15 elements. The number $C_m$ of ways to choose $m$ men from 10 men is the number of $m$-combinations of a set of 10 elements.
2. By product rule, the number of ways to form a commiitee with $w$ women and $m$ man is $C_{w} \cdot C_{m} = C(15, w) \cdot C(10, m))$.
3. By sum rule, the total number of ways to form a committee if it must have more women than men is

$C = C(15, 6)\cdot C(10, 0) + C(15, 5)\cdot C(10, 1)+C(15, 4)\cdot C(10, 2)$