(1a)
$26^2\cdot10^5$
(1b)
$26\cdot25\cdot10\cdot9\cdot8\cdot7\cdot6$
(2)
$6^4$
(3)
$20!$
(4a)
$4!=24$
$i_p$ is a sequence: person $p$ is assigned an instrument $i$. Some possible sequences are
- $1\stxt{John}$, $2\stxt{Jim}$, $3\stxt{Jay}$, $4\stxt{Jack}$
- $1\stxt{John}$, $2\stxt{Jim}$, $3\stxt{Jack}$, $4\stxt{Jay}$
- $1\stxt{John}$, $2\stxt{Jack}$, $3\stxt{Jay}$, $4\stxt{Jim}$
- $1\stxt{Jack}$, $2\stxt{Jim}$, $3\stxt{Jay}$, $4\stxt{John}$
We want all permutations because each person is different and each instrument is different.
(4b)
$2\cdot1\cdot2\cdot1=4$ possible sequences. Let 1 denote the piano and let 2 denote the drums. Then all possible sequences are
- $1\stxt{Jay}$, $2\stxt{Jack}$, $3\stxt{John}$, $4\stxt{Jim}$
- $1\stxt{Jay}$, $2\stxt{Jack}$, $3\stxt{Jim}$, $4\stxt{John}$
- $1\stxt{Jack}$, $2\stxt{Jay}$, $3\stxt{John}$, $4\stxt{Jim}$
- $1\stxt{Jack}$, $2\stxt{Jay}$, $3\stxt{Jim}$, $4\stxt{John}$
(5a)
$8\cdot2\cdot9=8\cdot18=80+64=144$
(5b)
$18$
(6)
$7^4$
(7a)
$6!$
(7b)
$2(3!)(3!)=2\cdot36=72$
(7c)
$4(3!)(3!)=4\cdot36=144$
(7d)
$6\cdot3\cdot2\cdot2\cdot1\cdot1=18\cdot4=72$
(8a)
$5!$
(8b)
$\frac{7!}{2!2!}$
(8c)
$\frac{11!}{4!4!2!}$
(8d)
$\frac{7!}{2!2!}$
(9)
$\frac{12!}{6!4!}$
(10a)
$8!$
(10b)
$6!2!7$
(10c)
$8\cdot4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=32\cdot36$
(10d)
$3!5!4$
(10e)
$4!2^4$
(11a)
$6!=30\cdot12\cdot2=720$
(11b)
$3!3!2!=6\cdot6\cdot2=72$
(11c)
$3!3!4$
(12a)
$30^5$
(12b)
$30\cdot29\cdot28\cdot27\cdot26=\frac{30!}{25!}$
(13)
There are (at least) two ways to look at this: “all at once” or “one at a time”.
The “all at once” approach is this: arbitrarily order the 20 persons, take the first person and have that first person shake hands with the other 19. Now remove that first person from the lot, since he/she has already shaken hands with everyone. Now have the second person shake hands with the other 18. So we have a total of 19+18 handshakes and we remove the second person from the lot. So the math looks like:
$19+18+17+…+3+2+1=\sum_{i=1}^{19}i=\frac{20\cdot19}{2}$
I call this approach “all at once” because we arbitrarily ordered the 20 persons in one shot, and then just went down the line.
Alternatively, we can do this “one at a time”: pick one person from the 20 and then pick a second person from the remaining 19. Then remove the permutations:
$\binom{20}{2}=\frac{20!}{18!2!}=\frac{20\cdot19}{2}$
Joe shaking hands with Tom is the same thing as Tom shaking hands with Joe. The permutations are irrelevant. So we can liken a handshake to a “committee of 2”. Then asking “how many handshakes take place” is the same as asking how many possible committees of 2 are there? And this is similar to example 4a.
(14)
This does not work: $F_1=52\cdot51\cdot50\cdot49\cdot48$. This reason is that the permutations are counted multiple times. $F_1$ is really the count of all possible permutations of the $52$ cards. For example:
$A_S,K_D,Q_H,J_C,10_S$
$K_D,A_S,Q_H,J_C,10_S$
These two permutations are the same hand in poker. But this same hand is counted multiple times in $F_1$. How many times? Every permutation of that hand is counted. How many permutations of that hand (actually of any hand) are there? $5!$. And we want to remove the multiple counts for every possible poker hand. So we divide $F_1$ by $5!$ and the correct answer is
$\frac{52\cdot51\cdot50\cdot49\cdot48}{5!}=\frac{52!}{47!5!}=\binom{52}{5}$
(15)
To choose the 5 men, we have $\binom{12}{5}$ possibilities. To choose the 5 women, we have $\binom{10}{5}$ possibilities. When we liken each choice of 5 to an experiment, the general principle of counting says that the total number of all possible choices for 5 men and 5 women is $\binom{12}{5}\binom{10}{5}$.
To pair the men and women, first arbitrarily fix the order of the 5 men. The first man can be paired with any of the 5 women. The second man can be paired with any of the 4 remaining women. And so on. And we get a count of $5!$.
Notice how this pairing is different from problem (7d). In that problem, we counted the number of ways that boys and girls can sit in a row if no two people of the same sex are allowed to sit together. This is similar to the problem of pairing men and women for dancing.
But there is an important difference. If we have 5 men and 5 women and we line them up in a row with alternating gender, we get a count of $2\wts5\wts5\wts4\wts4\wts3…=2\wts5!5!$. To get the possible pairings of the 5 and 5, we first divide this count by 2 because the male or female going first is irrelevant for pairing. We must also divide by $5!$ because the order of the pairs is irrelevant for purposes of pairing. So again the total count is $\frac{2\wts5!5!}{2\wts5!}=5!$.
And now we go back to the original question: we have a count of $\binom{12}{5}\binom{10}{5}$ to select the 5 men and 5 women. And for each selection of 5 and 5, we have a count of $5!$ ways to pair them up. Using the general principle of counting, we get a total of $\binom{12}{5}\binom{10}{5}\wt5!$.
(16a)
$\binom{6}{2}\binom{7}{0}\binom{4}{0}+\binom{6}{0}\binom{7}{2}\binom{4}{0}+\binom{6}{0}\binom{7}{0}\binom{4}{2}=15+21+6=42$
(16b)
$6\wt7+6\wt4+7\wt4=42+24+28=42+52=94$
(17)
Let’s look at two options:
$F_1=\binom{10}{7}=\frac{10!}{7!3!}$
or
$F_2=10\wt9\wt8\wt7\wt6\wt5\wt4=\frac{10!}{3!}$
$F_1$ excludes permutations while $F_2$ includes permutations. So the question is, does this question imply the need for permutations? Well let’s look at
$1,2,3,4,5,6,7$
$2,1,3,4,5,6,7$
Different permutations of the same combination. In the first permutation, child 1 will get gift 1 while child 2 will get gift 2. In the second permutation, child 2 will get gift 1 and child 1 will get gift 2. These are different results since the gifts are different. So we actually want $F_2=\frac{10!}{3!}$.
(18)
The general counting principle definitely applies appear since “for each selection of 2 R’s, for each selection of 2 D’s, we select 3 I’s” is the algorithm we want. Note that order doesn’t matter anywhere in these various selections: 2 D’s are 2 D’s regardless of order. Selecting I’s first changes nothing about the final committee:
$\binom{5}{2}\binom{6}{2}\binom{4}{3}=10\wt15\wt4=600$
(19a)
$\binom{8}{3}\bop\binom{4}{3}+\binom{2}{1}\binom{4}{2}\bcp$
(19b)
$\binom{6}{3}\bop\binom{6}{3}+\binom{2}{1}\binom{6}{2}\bcp$
(19c)
$\binom{7}{3}\binom{5}{3}+\binom{7}{2}\binom{5}{3}+\binom{7}{3}\binom{5}{2}$
(20a)
$\binom{6}{5}+\binom{2}{1}\binom{6}{4}$
(20b)
$\binom{6}{5}+\binom{6}{3}$
(21)
r’s represent right moves; u’s represent up moves. To get from A to B, we must do exactly 4 r’s and 3 u’s. So any path we take will require exactly 7 moves.
We definitely don’t want permutations of the r’s amongst themselves. Nor do we want permutations of the u’s amongst themselves. But we do want permutations between those two groups. That is:
rrrruuu
rrruruu
should be counted twice because each represents a different path. So, starting with $7!$, we want to remove the permutations within the 4 r’s and the permutations within the 3 u’s:
$\frac{7!}{4!3!}$
(22)
From A to C: $\frac{4!}{2!2!}=3\wt2=6$
From C to B: $\frac{3!}{2!1!}=3$
From A to B: $6\wt3=18$
(23)
$3!2^3$
(24)
$(3x^2+y)^5=\sum_{i=0}^{5}\binom{5}{i}(3x^2)^iy^{5-i}$
(25)
$\binom{52}{13,13,13,13}=\frac{52!}{13!13!13!13!}$
(27)
$\binom{12}{3,4,5}=\frac{12!}{3!4!5!}$
TD
- calc-1
- calc-2
- lin-alg-1
- lin-alg-2
- adv-calc-1
- adv-calc-2
- manifolds
- mtrx-calc
- prob-1
- prob-2
- math-stat
- stat-infer
- real-an-1
- stoch-proc
(28a)
We think of sequences.
$T=t_1,t_2,t_3,t_4,t_5,t_6,t_7,t_8$
Each teacher is assigned a school. So teacher #1 is assigned school $t_1$. We want to find the number of possible $T$’s. $T$ is really a sequence of 8 experiments so we can use the general counting principle.
Teacher #1 can be assigned to any of the 4 schools. So experiment #1 has 4 possibilities. Similarly, experiment #2 has 4 possibilities, etc.
The number of possible $T$’s is $4^8$.
(28b)
$\binom{8}{2,2,2,2}=\frac{8!}{2!2!2!2!}$
$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}=\frac{8\wt7}{2}\frac{6\wt5}{2}\frac{4\wt3}{2}\frac{2\wt1}{2}$
(29a)
Let $U$ denote a US lifter, $R$ denote a Russian lifter, $C$ denote a Chinese lifter, and $N$ denote a Canadian lifter. Then this question boils down to how many letter arrangements are there of 3 $U$’s, 4 $R$’s, 2 $C$’s, and 1 $N$.
Just as in the PEPPER example (3d), we take all permutations and divide by the dup permutations:
$\frac{10!}{4!3!2!}=\binom{10}{4,3,2,1}$
(29b)
Count of arrangements of non-US: $\frac{7!}{4!2!}$.
For each possible arrangement of non-US, there are $3$ slots on left to put a single $U$. That is, for each arrangement of non-US, you can place a $U$ in either slot $0$, slot $1$, or slot $2$.
And for each of these arrangements, there are $3$ slots on the right in which to put $2$ $U$’s: $\binom32$
Then by the general counting principle we get a count of $\frac{7!}{4!2!}3\binom32$.