Chapter 6: Permutations and Combinations

The word ROSE has 4 distinct letters: R, O, S, E. Since repetition is not allowed, the number of 4-letter words equals the number of permutations of 4 distinct letters taken all at a time.

$$\,{}^{4}P_4 = 4! = 4 \times 3 \times 2 \times 1 = 24$$

Each signal uses 2 distinct flags from the 4 available flags, arranged one below the other. Since the top flag and bottom flag are in different positions, the order matters; the number of signals is the number of permutations of 4 flags taken 2 at a time.

$$\,{}^{4}P_2 = \dfrac{4!}{(4-2)!} = \dfrac{4 \times 3 \times 2!}{2!} = 12$$

For a 2-digit number to be even, its units digit must be even. From the digits $$\{1, 2, 3, 4, 5\}$$ the even digits are $$2$$ and $$4$$, giving $$2$$ choices.

Since repetition is allowed, the tens digit can be any of the 5 given digits, giving $$5$$ choices.

By the fundamental principle of counting, the number of 2-digit even numbers is

$$5 \times 2 = 10$$

A signal can use $$2, 3, 4$$ or $$5$$ flags. Since the flags are arranged in a specific order on the staff, the number of signals of each length is a permutation.

Number of signals using exactly 2 flags: $$\,{}^{5}P_2 = 5 \times 4 = 20$$

Number of signals using exactly 3 flags: $$\,{}^{5}P_3 = 5 \times 4 \times 3 = 60$$

Number of signals using exactly 4 flags: $$\,{}^{5}P_4 = 5 \times 4 \times 3 \times 2 = 120$$

Number of signals using all 5 flags: $$\,{}^{5}P_5 = 5! = 120$$

By the addition principle, the total number of signals is

$$20 + 60 + 120 + 120 = 320$$

Recall that $$n! = n \times (n-1) \times (n-2) \times \cdots \times 1$$.

(i) $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

(ii) $$7! = 7 \times 6 \times 5! = 7 \times 6 \times 120 = 5040$$

(iii) $$7! - 5! = 5040 - 120 = 4920$$

(i) Write $$7! = 7 \times 6 \times 5!$$ to cancel:

$$\dfrac{7!}{5!} = \dfrac{7 \times 6 \times 5!}{5!} = 7 \times 6 = 42$$

(ii) Similarly, $$12! = 12 \times 11 \times 10!$$:

$$\dfrac{12!}{10!\,2!} = \dfrac{12 \times 11 \times 10!}{10! \times 2!} = \dfrac{12 \times 11}{2} = 66$$

Substitute $$n = 5$$ and $$r = 2$$:

$$\dfrac{n!}{r!\,(n-r)!} = \dfrac{5!}{2!\,(5-2)!} = \dfrac{5!}{2!\,3!}$$

$$= \dfrac{5 \times 4 \times 3!}{2! \times 3!} = \dfrac{5 \times 4}{2} = 10$$

Take a common denominator on the LHS using $$9! = 9 \times 8!$$:

$$\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{9}{9!} + \dfrac{1}{9!} = \dfrac{10}{9!}$$

Multiply numerator and denominator by $$10$$ to bring the denominator to $$10!$$:

$$\dfrac{10}{9!} = \dfrac{10 \times 10}{10 \times 9!} = \dfrac{100}{10!}$$

Comparing with $$\dfrac{x}{10!}$$, we get $$x = 100$$.

The word ALLAHABAD has 9 letters. Their frequencies are: A occurs 4 times, L occurs 2 times, and H, B, D each occur once.

The number of distinguishable permutations of $$n$$ objects in which $$p_1, p_2, \ldots$$ are alike is $$\dfrac{n!}{p_1!\,p_2!\,\cdots}$$. Hence

$$\dfrac{9!}{4!\,2!} = \dfrac{362880}{24 \times 2} = \dfrac{362880}{48} = 7560$$

A 4-digit number is formed by choosing 4 of the 9 digits $$1, 2, \ldots, 9$$ and arranging them in order. Since repetition is not allowed and (because $$0$$ is not among the available digits) the leading digit is automatically non-zero, the number of such arrangements is

$$\,{}^{9}P_4 = 9 \times 8 \times 7 \times 6 = 3024$$

Numbers between $$100$$ and $$1000$$ are exactly the 3-digit numbers. The hundreds digit cannot be $$0$$.

Hundreds digit: any non-zero digit from $$\{0, 1, 2, 3, 4, 5\}$$ — $$5$$ choices.

Tens digit: any of the remaining $$5$$ digits — $$5$$ choices.

Units digit: any of the remaining $$4$$ digits — $$4$$ choices.

By the multiplication principle, the required count is

$$5 \times 5 \times 4 = 100$$

Write both permutations using factorials:

$$5 \times \dfrac{4!}{(4-r)!} = 6 \times \dfrac{5!}{(5-(r-1))!}$$

$$\dfrac{5 \times 24}{(4-r)!} = \dfrac{6 \times 120}{(6-r)!}$$

$$\dfrac{120}{(4-r)!} = \dfrac{720}{(6-r)!}$$

Cross-multiply: $$(6-r)! = 6 \times (4-r)!$$. Since $$(6-r)! = (6-r)(5-r)(4-r)!$$,

$$(6-r)(5-r) = 6$$

$$r^2 - 11r + 30 = 6 \Rightarrow r^2 - 11r + 24 = 0$$

$$(r-3)(r-8) = 0 \Rightarrow r = 3 \text{ or } 8$$

For $$\,{}^{4}P_r$$ to be defined we need $$0 \le r \le 4$$, so $$r = 3$$.

There are $$4 + 3 + 2 = 9$$ discs in total, with 4 red (alike), 3 yellow (alike), and 2 green (alike).

The number of distinguishable arrangements of $$n$$ objects with $$p_1, p_2, \ldots$$ alike is $$\dfrac{n!}{p_1!\,p_2!\,\cdots}$$. Hence

$$\dfrac{9!}{4!\,3!\,2!} = \dfrac{362880}{24 \times 6 \times 2} = \dfrac{362880}{288} = 1260$$

Use the property: if $$\,{}^{n}C_x = \,{}^{n}C_y$$ with $$x \ne y$$, then $$x + y = n$$.

Here $$x = 9, y = 8$$, so $$n = 9 + 8 = 17$$.

Therefore $$\,{}^{n}C_{17} = \,{}^{17}C_{17} = 1$$.

The total number of people is $$2 + 3 = 5$$. Choosing any 3 of them:

$$\,{}^{5}C_3 = \dfrac{5!}{3!\,2!} = 10$$

Committees with 1 man and 2 women: choose 1 of the 2 men and 2 of the 3 women.

$$\,{}^{2}C_1 \times \,{}^{3}C_2 = 2 \times 3 = 6$$

The word INVOLUTE has 8 distinct letters: 4 vowels (I, O, U, E) and 4 consonants (N, V, L, T).

Choose 3 vowels from 4: $$\,{}^{4}C_3 = 4$$ ways.

Choose 2 consonants from 4: $$\,{}^{4}C_2 = 6$$ ways.

Each chosen group of 5 distinct letters can be arranged in $$5! = 120$$ ways.

Required number of words:

$$4 \times 6 \times 5! = 4 \times 6 \times 120 = 2880$$

AGAIN has 5 letters with A repeated twice. The total number of distinguishable arrangements is

$$\dfrac{5!}{2!} = \dfrac{120}{2} = 60$$

To find the $$50^{\text{th}}$$ word in dictionary order, list the letters alphabetically: A, A, G, I, N.

Words starting with A: fix A first; arrange the remaining A, G, I, N (all distinct) in $$4! = 24$$ ways. These occupy positions 1–24.

Words starting with G: arrange the remaining A, A, I, N in $$\dfrac{4!}{2!} = 12$$ ways. Positions 25–36.

Words starting with I: arrange A, A, G, N in $$\dfrac{4!}{2!} = 12$$ ways. Positions 37–48.

The next 12 words (positions 49–60) start with N. After fixing N, the remaining letters A, A, G, I are arranged in alphabetical order.

Words starting with NA: arrange A, G, I in $$3! = 6$$ ways. These occupy positions 49–54, listed as:

NAAGI, NAAIG, NAGAI, NAGIA, NAIAG, NAIGA.

So the $$49^{\text{th}}$$ word is NAAGI and the $$50^{\text{th}}$$ word is NAAIG.

To exceed $$1000000$$, the number must have 7 digits and its leading digit must be non-zero.

The available digits are 1, 2, 0, 2, 4, 2, 4 — seven digits in which 2 occurs 3 times, 4 occurs 2 times, and 0, 1 occur once each.

Total number of 7-digit arrangements (allowing 0 in the leading position) is

$$\dfrac{7!}{3!\,2!} = \dfrac{5040}{12} = 420$$

Arrangements with 0 as the leading digit: fix 0 first and arrange the remaining 6 digits 1, 2, 2, 2, 4, 4 in

$$\dfrac{6!}{3!\,2!} = \dfrac{720}{12} = 60$$ ways.

Required number = $$420 - 60 = 360$$.

First seat the 5 girls in a row. Since the girls are all distinct, this can be done in $$5! = 120$$ ways.

Once the girls are seated, they create 6 gaps in which boys may be placed — one gap before the first girl, one gap after the last girl, and one gap between each pair of adjacent girls.

To ensure that no two boys are together, each boy must be placed in a separate gap. The number of ways to place the 3 distinct boys into 6 distinct gaps (order matters, since the seats are distinct) is

$$\,{}^{6}P_3 = 6 \times 5 \times 4 = 120$$

By the multiplication principle, the required number of seating arrangements is

$$5! \times \,{}^{6}P_3 = 120 \times 120 = 14400$$

For a number to be even, its units digit must be even. Among the digits $$1, 2, 3, 4, 5, 6$$ the even digits are $$2, 4, 6$$ — that is $$3$$ choices for the units place.

Since repetition is allowed, the hundreds and tens places can each be filled by any of the 6 given digits:

Hundreds place: $$6$$ ways; tens place: $$6$$ ways; units place: $$3$$ ways.

By the fundamental principle of counting, the number of 3-digit even numbers is

$$6 \times 6 \times 3 = 108$$

A 4-letter code is an ordered arrangement of 4 letters chosen from the first 10 letters of the English alphabet (A to J), with no letter repeated.

This is the number of permutations of 10 distinct letters taken 4 at a time:

$$\,{}^{10}P_4 = \dfrac{10!}{(10-4)!} = \dfrac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$$

The telephone number has 5 digits, and the first two digits are fixed as 6 and 7. So only the last 3 places remain to be filled.

Since the number starts with 67 and no digit may be repeated, the digits 6 and 7 are already used. The remaining 3 places must be filled using the other $$10 - 2 = 8$$ digits, without repetition:

Third place: $$8$$ ways; fourth place: $$7$$ ways; fifth place: $$6$$ ways.

By the fundamental principle of counting, the number of such telephone numbers is

$$8 \times 7 \times 6 = 336$$

Each toss of a coin has $$2$$ possible outcomes — head (H) or tail (T). The three tosses are independent, so the outcome of one toss does not affect the others.

By the fundamental principle of counting, the number of possible outcomes when the coin is tossed 3 times is

$$2 \times 2 \times 2 = 8$$

(These outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.)

Each signal uses 2 of the 5 different flags, placed one below the other. Since the upper position and the lower position are distinct, the order of the two flags matters.

Hence the number of signals is the number of permutations of 5 flags taken 2 at a time:

$$\,{}^{5}P_2 = \dfrac{5!}{(5-2)!} = \dfrac{5!}{3!} = 5 \times 4 = 20$$

Evaluate the two sides separately.

Left-hand side: $$3! + 4! = 6 + 24 = 30$$.

Right-hand side: $$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$$.

Since $$30 \ne 5040$$, we conclude that $$3! + 4! \ne 7!$$.

Write $$8! = 8 \times 7 \times 6!$$ so that $$6!$$ cancels:

$$\dfrac{8!}{6! \times 2!} = \dfrac{8 \times 7 \times 6!}{6! \times 2!} = \dfrac{8 \times 7}{2!} = \dfrac{56}{2} = 28$$

We are given $$\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$$. Use $$7! = 7 \times 6!$$ to rewrite the first term with denominator $$7!$$:

$$\dfrac{1}{6!} = \dfrac{7}{7 \times 6!} = \dfrac{7}{7!}$$

So the left-hand side becomes

$$\dfrac{7}{7!} + \dfrac{1}{7!} = \dfrac{8}{7!}$$

Now write this with denominator $$8! = 8 \times 7!$$:

$$\dfrac{8}{7!} = \dfrac{8 \times 8}{8 \times 7!} = \dfrac{64}{8!}$$

Comparing with $$\dfrac{x}{8!}$$ gives $$x = 64$$.

A 3-digit number has a hundreds, a tens and a units place. The digits available are $$1, 2, \ldots, 9$$ — nine digits — and no digit is repeated. Since $$0$$ is not among the available digits, the hundreds place is automatically non-zero.

The required count is therefore the number of permutations of 9 digits taken 3 at a time:

$$\,{}^{9}P_3 = \dfrac{9!}{(9-3)!} = \dfrac{9!}{6!} = 9 \times 8 \times 7 = 504$$

A 4-digit number uses the digits $$0, 1, 2, \ldots, 9$$ with no repetition. The thousands place cannot be $$0$$, otherwise it would not be a 4-digit number.

Thousands place: any digit except $$0$$ — $$9$$ choices.

Hundreds place: any of the remaining $$9$$ digits (now $$0$$ is allowed) — $$9$$ choices.

Tens place: any of the remaining $$8$$ digits — $$8$$ choices.

Units place: any of the remaining $$7$$ digits — $$7$$ choices.

By the multiplication principle, the number of 4-digit numbers with no digit repeated is

$$9 \times 9 \times 8 \times 7 = 4536$$

The available digits are $$1, 2, 3, 4, 6, 7$$ — six digits — and no digit is repeated. For the number to be even, its units digit must be even.

The even digits available are $$2, 4, 6$$, so the units place has $$3$$ choices.

After fixing the units digit, $$5$$ digits remain for the other two places:

Hundreds place: $$5$$ choices; tens place: $$4$$ choices.

By the multiplication principle, the number of 3-digit even numbers is

$$3 \times 5 \times 4 = 60$$

The digits available are $$1, 2, 3, 4, 5$$ — five digits — with no repetition.

Total 4-digit numbers: this is the number of permutations of 5 digits taken 4 at a time:

$$\,{}^{5}P_4 = \dfrac{5!}{(5-4)!} = \dfrac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$$

Even 4-digit numbers: the units digit must be even, i.e. $$2$$ or $$4$$ — $$2$$ choices.

After fixing the units digit, the remaining 3 places are filled from the remaining 4 digits:

Thousands place: $$4$$ choices; hundreds place: $$3$$ choices; tens place: $$2$$ choices.

Number of even 4-digit numbers $$= 2 \times 4 \times 3 \times 2 = 48$$.

Choosing a chairman and a vice-chairman from 8 persons amounts to filling 2 distinct positions, where no person may hold both positions.

Chairman: any of the $$8$$ persons — $$8$$ ways.

Vice-chairman: any of the remaining $$7$$ persons — $$7$$ ways.

Since the two positions are distinct, the required number is the number of permutations of 8 persons taken 2 at a time:

$$\,{}^{8}P_2 = 8 \times 7 = 56$$

Write each permutation using $$\,{}^{n}P_r = \dfrac{n!}{(n-r)!}$$:

$$\,{}^{n-1}P_3 = \dfrac{(n-1)!}{(n-1-3)!} = \dfrac{(n-1)!}{(n-4)!}$$

$$\,{}^{n}P_4 = \dfrac{n!}{(n-4)!}$$

Form the ratio:

$$\dfrac{\,{}^{n-1}P_3}{\,{}^{n}P_4} = \dfrac{(n-1)!}{(n-4)!} \times \dfrac{(n-4)!}{n!} = \dfrac{(n-1)!}{n!} = \dfrac{1}{n}$$

The given condition is $$\dfrac{\,{}^{n-1}P_3}{\,{}^{n}P_4} = \dfrac{1}{9}$$, so

$$\dfrac{1}{n} = \dfrac{1}{9} \Rightarrow n = 9$$

The word EQUATION has 8 letters — E, Q, U, A, T, I, O, N — all of which are distinct.

Using each letter exactly once, the number of words is the number of permutations of 8 distinct letters taken all at a time:

$$\,{}^{8}P_8 = 8! = 40320$$

The word MISSISSIPPI has 11 letters. Their frequencies are: I occurs 4 times, S occurs 4 times, P occurs 2 times, and M occurs once.

Total distinct permutations:

$$\dfrac{11!}{4!\,4!\,2!} = \dfrac{39916800}{24 \times 24 \times 2} = \dfrac{39916800}{1152} = 34650$$

Permutations with the four I's together: treat the four I's as a single block. We then arrange this block together with M, S, S, S, S, P, P — that is $$8$$ objects, in which S occurs 4 times and P occurs 2 times:

$$\dfrac{8!}{4!\,2!} = \dfrac{40320}{24 \times 2} = \dfrac{40320}{48} = 840$$

Permutations in which the four I's do NOT come together:

$$34650 - 840 = 33810$$

Use the property: if $$\,{}^{n}C_a = \,{}^{n}C_b$$ and $$a \ne b$$, then $$a + b = n$$.

Here $$\,{}^{n}C_8 = \,{}^{n}C_2$$ with $$8 \ne 2$$, so

$$n = 8 + 2 = 10$$

Therefore

$$\,{}^{n}C_2 = \,{}^{10}C_2 = \dfrac{10!}{2!\,8!} = \dfrac{10 \times 9}{2} = 45$$

A chord of a circle is a line segment joining 2 points on the circle. Through 21 given points, each chord is determined uniquely by a choice of 2 of the points, and the order of the two points does not matter.

Hence the number of chords is the number of combinations of 21 points taken 2 at a time:

$$\,{}^{21}C_2 = \dfrac{21!}{2!\,19!} = \dfrac{21 \times 20}{2} = 210$$

We must select 3 boys from 5 boys and 3 girls from 4 girls. Since this is a selection (the order does not matter), we use combinations.

Number of ways to choose 3 boys from 5: $$\,{}^{5}C_3 = \dfrac{5 \times 4 \times 3}{3!} = 10$$.

Number of ways to choose 3 girls from 4: $$\,{}^{4}C_3 = \dfrac{4!}{3!\,1!} = 4$$.

By the multiplication principle, the number of ways to form the team is

$$\,{}^{5}C_3 \times \,{}^{4}C_3 = 10 \times 4 = 40$$

Each selection must contain exactly 3 balls of each colour. The balls of each colour are chosen independently of the others.

Choose 3 red balls from 6: $$\,{}^{6}C_3 = \dfrac{6 \times 5 \times 4}{3!} = 20$$.

Choose 3 white balls from 5: $$\,{}^{5}C_3 = \dfrac{5 \times 4 \times 3}{3!} = 10$$.

Choose 3 blue balls from 5: $$\,{}^{5}C_3 = 10$$.

By the multiplication principle, the number of ways of selecting the 9 balls is

$$\,{}^{6}C_3 \times \,{}^{5}C_3 \times \,{}^{5}C_3 = 20 \times 10 \times 10 = 2000$$

A deck of 52 cards contains 4 aces and $$52 - 4 = 48$$ non-ace cards. Each combination of 5 cards must contain exactly one ace.

Choose 1 ace from the 4 aces: $$\,{}^{4}C_1 = 4$$ ways.

Choose the remaining 4 cards from the 48 non-ace cards:

$$\,{}^{48}C_4 = \dfrac{48 \times 47 \times 46 \times 45}{4!} = \dfrac{4669920}{24} = 194580$$

By the multiplication principle, the required number of 5-card combinations is

$$\,{}^{4}C_1 \times \,{}^{48}C_4 = 4 \times 194580 = 778320$$

Among the 17 players, 5 can bowl and the remaining $$17 - 5 = 12$$ cannot. The team of 11 must include exactly 4 bowlers, so the other $$11 - 4 = 7$$ players are chosen from the 12 non-bowlers.

Choose 4 bowlers from 5: $$\,{}^{5}C_4 = 5$$ ways.

Choose 7 non-bowlers from 12: $$\,{}^{12}C_7 = \,{}^{12}C_5 = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5!} = \dfrac{95040}{120} = 792$$ ways.

By the multiplication principle, the number of ways to select the team is

$$\,{}^{5}C_4 \times \,{}^{12}C_7 = 5 \times 792 = 3960$$

The bag has 5 black balls and 6 red balls. We must select 2 black and 3 red balls; the order of selection does not matter, so we use combinations.

Choose 2 black balls from 5: $$\,{}^{5}C_2 = \dfrac{5 \times 4}{2!} = 10$$.

Choose 3 red balls from 6: $$\,{}^{6}C_3 = \dfrac{6 \times 5 \times 4}{3!} = 20$$.

By the multiplication principle, the number of ways of selecting the 5 balls is

$$\,{}^{5}C_2 \times \,{}^{6}C_3 = 10 \times 20 = 200$$

The student must choose 5 courses out of 9, but 2 specific courses are compulsory and so are always included in the programme.

Since those 2 courses are already chosen, the student only needs to choose the remaining $$5 - 2 = 3$$ courses from the remaining $$9 - 2 = 7$$ courses.

$$\,{}^{7}C_3 = \dfrac{7 \times 6 \times 5}{3!} = \dfrac{210}{6} = 35$$

The word DAUGHTER has 8 letters. The vowels are A, U, E — $$3$$ vowels — and the consonants are D, G, H, T, R — $$5$$ consonants. All letters are distinct.

Choose 2 vowels from 3: $$\,{}^{3}C_2 = 3$$ ways.

Choose 3 consonants from 5: $$\,{}^{5}C_3 = \dfrac{5 \times 4 \times 3}{3!} = 10$$ ways.

Each such selection consists of $$2 + 3 = 5$$ distinct letters, which can be arranged among themselves to form words in $$5! = 120$$ ways.

By the multiplication principle, the required number of words is

$$\,{}^{3}C_2 \times \,{}^{5}C_3 \times 5! = 3 \times 10 \times 120 = 3600$$

The word EQUATION has 8 distinct letters. The vowels are E, U, A, I, O — $$5$$ vowels — and the consonants are Q, T, N — $$3$$ consonants.

We require all the vowels to occur together and all the consonants to occur together. Treat the group of 5 vowels as one block and the group of 3 consonants as another block.

These 2 blocks can be arranged in $$2! = 2$$ ways.

Within the vowel block, the 5 distinct vowels can be arranged in $$5! = 120$$ ways.

Within the consonant block, the 3 distinct consonants can be arranged in $$3! = 6$$ ways.

By the multiplication principle, the required number of words is

$$2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440$$

The word EXAMINATION has 11 letters. Their frequencies are: A occurs 2 times, I occurs 2 times, N occurs 2 times, and E, X, M, T, O each occur once.

In dictionary order, the letters arranged alphabetically are A, A, E, I, I, M, N, N, O, T, X. The smallest letter is A and the next is E.

So every word that comes before the first word starting with E must start with A. We count the words that start with A.

Fix one A in the first position. The remaining 10 letters to be arranged are A, E, X, M, I, N, I, T, O, N — among these, I occurs 2 times and N occurs 2 times (A now occurs only once).

Number of such arrangements:

$$\dfrac{10!}{2!\,2!} = \dfrac{3628800}{4} = 907200$$

Hence there are $$907200$$ words before the first word starting with E.

A number is divisible by 10 if and only if its units digit is 0. So in every required number the units place must be occupied by 0.

The number has 6 digits and uses the digits $$0, 1, 3, 5, 7, 9$$ with no repetition. Fixing 0 in the units place uses up the digit 0.

The remaining 5 places are then filled by the remaining digits $$1, 3, 5, 7, 9$$ (all non-zero, so there is no leading-zero problem), with no repetition:

$$5! = 120$$

Hence $$120$$ such 6-digit numbers can be formed.

Each word must contain 2 different vowels and 2 different consonants, making 4 distinct letters in all.

Choose 2 vowels from the 5 available vowels: $$\,{}^{5}C_2 = \dfrac{5 \times 4}{2!} = 10$$ ways.

Choose 2 consonants from the 21 available consonants: $$\,{}^{21}C_2 = \dfrac{21 \times 20}{2!} = 210$$ ways.

Each selection of 4 distinct letters can be arranged to form words in $$4! = 24$$ ways.

By the multiplication principle, the required number of words is

$$\,{}^{5}C_2 \times \,{}^{21}C_2 \times 4! = 10 \times 210 \times 24 = 50400$$

Part I has 5 questions and Part II has 7 questions. A student attempts 8 questions in all, with at least 3 from each part.

Let the student attempt $$a$$ questions from Part I and $$b$$ from Part II, where $$a + b = 8$$ with $$a \ge 3$$, $$b \ge 3$$, $$a \le 5$$, $$b \le 7$$. The possible pairs $$(a, b)$$ are $$(3, 5)$$, $$(4, 4)$$ and $$(5, 3)$$.

3 from Part I, 5 from Part II: $$\,{}^{5}C_3 \times \,{}^{7}C_5 = 10 \times 21 = 210$$

4 from Part I, 4 from Part II: $$\,{}^{5}C_4 \times \,{}^{7}C_4 = 5 \times 35 = 175$$

5 from Part I, 3 from Part II: $$\,{}^{5}C_5 \times \,{}^{7}C_3 = 1 \times 35 = 35$$

By the addition principle, the required number of ways is

$$210 + 175 + 35 = 420$$

A deck of 52 cards contains 4 kings and $$52 - 4 = 48$$ non-king cards. Each combination of 5 cards must contain exactly one king.

Choose 1 king from the 4 kings: $$\,{}^{4}C_1 = 4$$ ways.

Choose the remaining 4 cards from the 48 non-king cards:

$$\,{}^{48}C_4 = \dfrac{48 \times 47 \times 46 \times 45}{4!} = \dfrac{4669920}{24} = 194580$$

By the multiplication principle, the required number of 5-card combinations is

$$\,{}^{4}C_1 \times \,{}^{48}C_4 = 4 \times 194580 = 778320$$

There are $$5 + 4 = 9$$ persons to be seated in a row of 9 positions, numbered 1 to 9.

The even-numbered positions are 2, 4, 6, 8 — that is exactly $$4$$ positions. The 4 women must occupy these 4 even places.

The 4 women can be arranged in the 4 even places in $$4! = 24$$ ways.

The remaining 5 positions (the odd places 1, 3, 5, 7, 9) are occupied by the 5 men, who can be arranged in $$5! = 120$$ ways.

By the multiplication principle, the number of such arrangements is

$$4! \times 5! = 24 \times 120 = 2880$$

From a class of 25 students, 10 are to be chosen. There are 3 special students who will either all join or all stay out. Consider the two cases.

Case 1: all 3 special students join. Then $$10 - 3 = 7$$ more students must be chosen from the remaining $$25 - 3 = 22$$ students:

$$\,{}^{22}C_7 = 170544$$

Case 2: none of the 3 special students joins. Then all 10 students must be chosen from the remaining 22 students:

$$\,{}^{22}C_{10} = 646646$$

By the addition principle, the number of ways to choose the excursion party is

$$\,{}^{22}C_7 + \,{}^{22}C_{10} = 170544 + 646646 = 817190$$

The word ASSASSINATION has 13 letters. Their frequencies are: A occurs 3 times, S occurs 4 times, I occurs 2 times, N occurs 2 times, and T, O each occur once. (Check: $$3 + 4 + 2 + 2 + 1 + 1 = 13$$.)

We require all 4 S's to be together. Treat the four S's as a single block. We then arrange this block together with the remaining 9 letters A, A, A, I, I, N, N, T, O — that is $$9 + 1 = 10$$ objects.

Among these 10 objects, A occurs 3 times, I occurs 2 times, N occurs 2 times, while the S-block, T and O occur once each. The number of arrangements is

$$\dfrac{10!}{3!\,2!\,2!} = \dfrac{3628800}{6 \times 2 \times 2} = \dfrac{3628800}{24} = 151200$$

Since the four S's inside the block are identical, the block can be arranged internally in only $$1$$ way, so it contributes no extra factor.

Hence the required number of arrangements is $$151200$$.