Permutations and Combinations — Practice

Question 1

How many arrangements of the letters of ARRANGE have no two $R$s together?

Accepted Answer

The non-$R$ letters are $A,A,N,G,E$: $\dfrac{5!}{2!}=60$ arrangements.

They create six gaps; choose two for the $R$s: $\binom{6}{2}=15$.

$60	imes15=900$.

Check: total arrangements are $\dfrac{7!}{2!\,2!}=1260$, and those with the $R$s together are $\dfrac{6!}{2!}=360$; $1260-360=900$.

Question 2

A team of $4$ is chosen from $7$ boys and $5$ girls, with at least one girl. In how many ways?

Accepted Answer

Total teams: $\binom{12}{4}=495$.

All boys: $\binom{7}{4}=35$.

$495-35=460$.

Reserving a girl first would give $5	imes\binom{11}{3}=825$, which counts a team with three girls three times over.

Question 3

How many selections of three letters can be made from the letters of the word PROPORTION?

Accepted Answer

Letters: $O^3$, $P^2$, $R^2$, $T$, $I$, $N$ — six distinct letters.

All different: $\binom{6}{3}=20$.

One pair plus one other: the pair is $O$, $P$ or $R$, so $3	imes5=15$.

All three alike: only $O$, so $1$.

Total $20+15+1=36$.

Question 4

Find the number of positive integer solutions of $x+y+z+w=20$ with each variable at least $2$.

Accepted Answer

Put $x'=x-2$ and similarly for the others, all non-negative, with $x'+y'+z'+w'=12$.

$\binom{12+3}{3}=\binom{15}{3}=455$.

Question 5

In how many ways can $10$ identical pens be distributed among $4$ children so that each gets at least one, and no child gets more than $4$?

Accepted Answer

After giving one to each, distribute $6$ among $4$ with each getting at most $3$.

Unrestricted: $\binom{9}{3}=84$.

Some child gets at least $4$: substitute to get $\binom{5}{3}=10$ each, so $40$.

Two children at least $4$ each needs $8>6$, impossible.

$84-40=44$.

Question 6

In how many ways can $12$ different books be divided equally among three students, and in how many ways into three unlabelled bundles?

Accepted Answer

Among three students: $\dfrac{12!}{4!\,4!\,4!}=34650$.

Into three unlabelled bundles: divide by $3!$, giving $\dfrac{34650}{6}=5775$.

The factor of six is exactly the number of ways the three bundles could be matched to the three students.

Question 7

Eight people, including a married couple, are seated around a round table. In how many arrangements are the couple not adjacent?

Accepted Answer

Total circular arrangements: $7!=5040$.

Adjacent: treat the couple as one unit, giving $6!$ circular arrangements, times $2$ for their internal order: $720	imes2=1440$.

$5040-1440=3600$.

Question 8

Six letters are put at random into six addressed envelopes. In how many ways does no letter reach its correct envelope?

Accepted Answer

$D_6=5\left(D_5+D_4ight)=5(44+9)=265$.

The series formula agrees: $720\left(1-1+	frac12-	frac16+	frac1{24}-	frac1{120}+	frac1{720}ight)=265$.

Question 9

Find the rank of the word RANK among all arrangements of its letters in dictionary order.

Accepted Answer

$R$: three smaller letters available, $3	imes3!=18$.

$A$: none smaller, $0$.

$N$: $K$ is smaller, $1	imes1!=1$.

$K$: none smaller, $0$.

Rank $=18+1+1=20$.

Question 10

Ten points lie in a plane, four of which are collinear and no other three are. How many triangles and how many straight lines do they determine?

Accepted Answer

Triangles: $\binom{10}{3}-\binom{4}{3}=120-4=116$.

Lines: $\binom{10}{2}-\binom{4}{2}+1=45-6+1=40$.

Question 11

How many five-digit numbers with distinct digits are divisible by $25$?

Accepted Answer

Endings $25$, $50$, $75$ are possible; $00$ repeats a digit.

Ending $50$: the first three positions use any of the remaining eight digits, $8	imes7	imes6=336$.

Endings $25$ and $75$: the leading digit cannot be $0$, so $7	imes7	imes6=294$ each.

$336+294+294=924$.

Question 12

Find the sum of all four-digit numbers formed using the digits $2,3,5,7$ without repetition.

Accepted Answer

Each digit occupies each place in $3!=6$ of the $24$ numbers.

Digit sum $=2+3+5+7=17$, so each place contributes $6	imes17=102$.

Sum $=102	imes1111=113322$.

Practice — Permutations and Combinations

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