O número de anagramas da palavra AMOR é
(A) 12 (B) 16 (C) 20 (D) 24 (E) 32

The word MATHEMATICS consists of 11 letters. The number of distinct ways to rearrange these letters is
(A) $11 !$
(B) $\frac { 11 ! } { 3 }$
(C) $\frac { 11 ! } { 6 }$
(D) $\frac { 11 ! } { 8 }$

From the numbers $1,2,3,4,5,6$, five numbers are selected with repetition allowed to satisfy the following conditions, and then all five-digit natural numbers that can be formed by arranging them in a line are counted. What is the number of such natural numbers? [4 points] (가) Each odd number is either not selected or selected exactly once. (나) Each even number is either not selected or selected exactly twice.
(1) 450
(2) 445
(3) 440
(4) 435
(5) 430

The number of ways to arrange all 5 letters $x, x, y, y, z$ in a row is? [2 points]
(1) 10
(2) 20
(3) 30
(4) 40
(5) 50

8. How many anagrams, even without meaning, are there of the word ``STUDIARE''? In how many of these anagrams can the word ``ARTE'' be read consecutively, as for example in ``SUARTEDI''?
How many anagrams, even without meaning, are there of the word ``VACANZA''? ``Mathematics knows no races or geographical boundaries; for mathematics, the cultural world is a single nation''
\footnotetext{Maximum duration of the exam: 6 hours. The use of scientific or graphical calculators is permitted provided they are not equipped with symbolic algebraic processing capability and do not have Internet connectivity. The use of a bilingual dictionary (Italian–language of the country of origin) is permitted for candidates whose native language is not Italian. It is not permitted to leave the Institute before 3 hours have elapsed from the distribution of the exam. }

5. The number of arrangements of the letters of the word BANANA in which the two N' s do not appear adjacently is

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(A) 40
(B) 60
(C) 80
(D) 100

Consider all possible permutations of the letters of the word ENDEANOEL.
Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.
Column I
(A) The number of permutations containing the word ENDEA is
(B) The number of permutations in which the letter E occurs in the first and the last positions is
(C) The number of permutations in which none of the letters D, L, N occurs in the last five positions is
(D) The number of permutations in which the letters A, E, O occur only in odd positions is
Column II
(p) $5 !$
(q) $2 \times 5 !$
(r) $7 \times 5 !$
(s) $21 \times 5 !$

Words of length 10 are formed using the letters $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}, \mathrm{G}, \mathrm{H}, \mathrm{I}, \mathrm{J}$. Let $x$ be the number of such words where no letter is repeated; and let $y$ be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{y}{9x} =$

All possible numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is
(1) 175
(2) 162
(3) 180
(4) 160

Total number of 6-digit numbers in which only and all the five digits $1, 3, 5, 7$ and 9 appears, is
(1) $\frac { 1 } { 2 } (6!)$
(2) $6!$
(3) $5 ^ { 6 }$
(4) $\frac { 5 } { 2 } (6!)$

The sum of all the 4-digit distinct numbers that can be formed with the digits $1, 2, 2$ and 3 is:
(1) 26664
(2) 122664
(3) 122234
(4) 22264

If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $C$ and $S$ do not come together, is $( 6 ! ) k$ then $k$ is equal to
(1) 2835
(2) 5670
(3) 1890
(4) 945

The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is $\_\_\_\_$.

The number of seven digits odd numbers, that can be formed using all the seven digits $1, 2, 2, 2, 3, 3, 5$ is

The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is
(1) 16800
(2) 33600
(3) 18000
(4) 14800

The number of words, which can be formed using all the letters of the word ``DAUGHTER'', so that all the vowels never come together, is
(1) 36000
(2) 37000
(3) 34000
(4) 35000

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is \_\_\_\_ .

Number of 4 letters words with or without meaning formed from the letters of the word PQRSSSTTUVV is\ (A) 1232\ (B) 1422\ (C) 1400\ (D) 1162

Consider the permutations of the eight letters of the word ``POSITION''.
(1) The number of permutations in which the two I's are adjacent and the two O's also are adjacent is $\mathbf { A B C }$.
(2) The number of permutations such that the permutations both begin and end with the letter I and furthermore the two O's are adjacent is $\mathbf{DEF}$.
(3) The number of permutations that both begin and end with the letter I is $\mathbf{GHI}$.
(4) The number of permutations of the 4 letters I, I, O, O is $\mathbf { J }$. Also, the number of permutations of the 4 letters $\mathrm { N } , \mathrm { P } , \mathrm { S } , \mathrm { T }$ is $\mathbf { K L }$.
Hence the number of permutations of POSITION which begin or end with either I or O, and furthermore in which none of letters $\mathrm { N } , \mathrm { P } , \mathrm { S } , \mathrm { T }$ are adjacent to each other is $\mathbf{MNO}$.