2 Marks Questions - Maths STD 11 Science Questions - Vidyadip

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33 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

Answer

Here total letters are 13 in the word ASSASSINATION in which A appears 3 times, S appears 4 times, 1 appears 2 times and N appears 2 times. Now four S's taken together become a single letter and other remaining letters taken with this single letter.
$\therefore $ Number of arrangements $ = \frac{{10!}}{{3!2!2!}} = \frac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}}{{3!2 \times 1 \times 2 \times 1}}$
$ \Rightarrow 10 \times 9 \times 8 \times 7 \times 6 \times 5 = 151200$

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Question 22 Marks

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer

There are 9 courses and number of courses to be selected are 5 in which 2 specific courses are compulsory.
We have to select 3 courses out of remaining 7 courses.
$\therefore $ Number of ways of selection ${ = ^7}{C_3}$
$\frac{{7!}}{{3!4!}} = \frac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}} = 35$

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Question 32 Marks

A bag contains 5 black balls and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot.

Answer

There are 5 black and 6 red balls. We have to select 2 black balls out of 5 black balls and 3 red balls out of 6 red balls.
$\therefore $ Number of ways of selection ${ = ^5}{C_2}{ \times ^6}{C_3}$
$= \frac{{5!}}{{2!3!}} \times \frac{{6!}}{{3!3!}} = 10 \times 20 = 200$

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Question 42 Marks

In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer

There are 5 bowlers and 12 other players in a team of 17 players we have to select 4 bowlers out of 5 bowlers and 7 other players out of 12 other players.
$\therefore $ Number of ways of selection ${ = ^5}{C_4}{ \times ^{12}}{C_7}$
$$
$=5{\times}792=3960$
$$

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Question 52 Marks

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Answer

There are 4 aces and 48 other cards in a deck of 52 cards. We have to select 1 ace out of 4 aces and 4 other cards out of 48 other cards.
$\therefore $ Number of ways of selection $=^4{C_1}{ \times ^{48}}{C_4}$
$= 4 \times \frac{{48 \times 47 \times 46 \times 45 }}{{4 \times 3 \times 2 \times 1}}$$= 778320$

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Question 62 Marks

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer

There are 5 boys and 4 girls. We have to select 3 out of 5 boys and 3 out of 4 girls.
$\therefore $ Number of ways of selection $=^5{C_{_3}}{ \times ^4}{C_{_3}}$
$= \frac{{5!}}{{3!2!}} \times \frac{{4!}}{{3!2!}} = 10 \times 4 = 40$

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Question 72 Marks

How many chords can be drawn through 21 points on a circle?

Answer

There are 21 points on the circumference on a circle. Since one and only one chord can be drawn by joining 2 distinct points, the required number of chords is given by
${21_{{C_2}}} = \frac{{21!}}{{2!19!}} = \frac{{21 \times 20 \times 19!}}{{2 \times 19!}} = 210$

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Question 82 Marks

Determine n if $^{2n}{C_3}{\ : \ ^n}{C_3} = 11:1$

Answer

Here,we have $^{2n}{C_3}{\ : \ ^n}{C_3} = 11:1$
$\Rightarrow \frac{2 \mathrm{n}_{\mathrm{C}_{3}}}{\mathrm{n}_{\mathrm{C}_{3}}}=\frac{11}{1}$
$ \Rightarrow \frac{{(2n)!}}{{3!(2n - 3)!}} \times \frac{{3!(n - 3)!}}{{n!}} = \frac{{11}}{1}$
$ \Rightarrow \frac{{(2n)(2n - 1)(2n - 2)(2n - 3)!}}{{3!(2n - 3)!}}$$ \times \frac{{3!(n - 3)!}}{{n(n - 1)(n - 2)(n - 3)!}} = \frac{{11}}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2)}{3 !}}{\frac{n \times(n-1) \times(n-2)}{3 !}}=\frac{11}{1}$
$ \Rightarrow \frac{{(2n)(2n - 1)(2n - 2)}}{{n(n - 1)(n - 2)}} = \frac{{11}}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times 2 \times(n-1)}{n \times(n-1) \times(n-2)}=\frac{11}{1}$
$\Rightarrow \frac{4 \times n \times(2 n-1)}{n \times(n-2)}=\frac{11}{1}$
$ \Rightarrow \frac{{4(2n - 1)}}{{n - 2}} = \frac{{11}}{1}$
$ \Rightarrow $ $4 \times(2 n-1)=11 \times(n-2)$
$ \Rightarrow 8n - 4 = 11n - 22$
$ \Rightarrow $ 3n = 18
$ \therefore$ n = 6

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Question 92 Marks

Determine n if $^{2n}{C_3}{:^n}{C_2} = 12:1$

Answer

Here,we have $^{2 n} C_{3}:^{n} C_{3}=12: 1$
$\Rightarrow \frac{2 \mathrm{n}_{\mathrm{C}_{3}}}{\mathrm{n}_{\mathrm{C}_{3}}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n !}{3 !(2 n-3) !}}{\frac{n !}{3 !(n-3) !}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2) \times(2 n-3) !}{3 !(2 n-3) !}}{\frac{n \times(n-1) \times(n-2) \times(n-3) !}{3 !(n-3) !}}=\frac{12}{1}$
$\Rightarrow \frac{\frac{2 n \times(2 n-1) \times(2 n-2)}{3 !}}{\frac{n \times(n-1) \times(n-2)}{3 !}}=\frac{12}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times(2 n-2)}{n \times(n-1) \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{2 n \times(2 n-1) \times 2 \times(n-1)}{n \times(n-1) \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{4 \times n \times(2 n-1)}{n \times(n-2)}=\frac{12}{1}$
$\Rightarrow \frac{4 \times(2 n-1)}{(n-2)}=\frac{12}{1}$
$\Rightarrow 4 \times(2 n-1)=12 \times(n-2)$
$\Rightarrow$ 8 n - 4 = 12 n - 24
$\Rightarrow$ 12 n - 8 n = 24 - 4
$\Rightarrow$ 4 n = 20
$\therefore$ n = 5

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Question 102 Marks

How many words with or without, meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?

Answer

Total number of letters in word EQUATION $= 8$
Number of letters used $($all different$) = 8$
$\therefore $ Number of permutation $=\ ^8P_8$
$= \frac{{8!}}{{0!}} = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

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Question 112 Marks

Find n if $^{n - 1}{P_3}{:^n}{P_4} = 1:9$

Answer

Here $^{n - 1}{P_3}{:^n}{P_4} = 1:9$
$\therefore \frac{{(n - 1)!}}{{(n - 4)!}}:\frac{{n!}}{{(n - 4)!}} = 1:9$.
$ \Rightarrow \frac{{(n - 1)!}}{{(n - 4)!}} \times \frac{{(n - 4)!}}{{n!}} = \frac{1}{9}$
$ \Rightarrow \frac{{(n - 1)!}}{{n!}} = \frac{1}{9} \Rightarrow \frac{{(n - 1)!}}{{n(n - 1)!}} = \frac{1}{9}$
$ \Rightarrow \frac{1}{n} = \frac{1}{9} \Rightarrow n = 9$

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Question 122 Marks

From a committee of 8 persons in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Answer

Here total number of persons = 8
Number of persons used (no persons is repeated) = 2
$\therefore $ Number of permutation ${ = ^8}{P_2}$
$= \frac{{8!}}{{6!}} = \frac{{8 \times 7 \times 6!}}{{6!}} = 56$

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Question 132 Marks

Find the number of $4-$digit numbers that can be formed using the digits $1, 2, 3, 4, 5$ if no digit is repeated. How many of these will be even?

Answer

Here total number of digits $= 5$
Number of digits used $($no digit is repeated$) = 4$
\therefore Number of permutations $=\ ^5P_4$
$= \frac{{5!}}{{1!}} = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Now the unit's place can be filled with any one of the digits $2, 4$ for even number.
\therefore Number of permutations $=\ ^2P_1 = 2$
Now the remaining three places can be filled with the remaining $4$ digits.
\therefore Number of permutation $=\ ^4P_3$
$\frac{{4!}}{{1!}} = 4 \times 3 \times 2 \times 1 = 24$
Hence total number of permutations of even numbers $= 2 \times 24 = 48$

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Question 142 Marks

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Answer

We know that, even number means that the last digit should be even,
No. of possible digits at one’s place = 3 (2, 4 and 6)
$\Rightarrow$ No. of permutations = $_{1}^{3} \mathrm{P}=\frac{3 !}{(3-1) !}$ = 3
One of a digit is taken at one’s place, Number of possible digits available = 5
$\Rightarrow$ No. of permutations = $_{2}^{5} P=\frac{5 !}{(5-2) !}=\frac{5 \times 4 \times 3 !}{3 !}$ = 20
Thus, a total number of permutations = 3 $\times$ 20 = 60

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Question 152 Marks

How many 4-digit numbers are there with no digit repeated?

Answer

Here total number of digits = 10
Number of digits used (no digit is repeated) = 4
Since 0 cannot be filled in the fourth place, so number of permutations for fourth place = 9
Now the remaining three places can be filled with 9 digits.
$\therefore $ Number of permutations = $^9{P_3}$
$= \frac{{9!}}{{6!}} = \frac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 504$
Hence total number of permutations $= 9 \times 504 = 4536$

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Question 162 Marks

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Answer

Here total number of digits = 9
Number of digits used (no digit is repeated) = 3
$\therefore $ Number of permutations = $^9{P_3}$
$= \frac{{9!}}{{6!}} = \frac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 504$

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Question 172 Marks

How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?

Answer

According to the problem, 2-digits i.e. 6 and 7 are fixed. Thus, 10 - 2 = 8-digits can be used in constructing the telephone numbers. There are 8-digits 0, 1, 2, 3, 4, 5, 8, 9.
The first number can be selected in 8 ways. After the selection of the first digit, we have 8 - 1 = 7-digits in hand, second digit can be selected in 7 ways and the third digit can be selected in 6 ways.
According to the fundamental principle of multiplication (FPM), the number of ways of selecting a digit for remaining three places:
= 8 $\times$ 7 $\times$ 6 = 336 ways

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Question 182 Marks

How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Answer

Here the unit place can be filled by any one of the first 10 letters of the English alphabet. So the unit place can be filled in 10 ways. The tens place can be filled in 9 ways by remaining 9 letters of the English alphabet. The hundreds place can be filled in 8 ways by remaining 8 letters of the English alphabet.
The thousands place can be filled in 7 ways by the remaining 7 letters of the English alphabet.
$\therefore $ Total number of 4-letter code numbers $ = 7 \times 8 \times 9 \times 10 = 5040$.

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Question 192 Marks

How many 3-digit even numbers can be formed, from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Answer

Here the unit place can be filled by any one of the digits 2, 4, 6. So the unit place can be filled in 3 ways. Now the tens and hundreds place can be filled by any one of the digits 1, 2, 3, 4, 5, 6. So the tens and hundreds place can be filled in 6 ways each.
$\therefore $ Total number of 3-digit even numbers $ = 6 \times 6 \times 3 = 108$

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Question 202 Marks

Find the number of permutations of the letters of the word ALLAHABAD.

Answer

The word ‘ALLAHABAD’ has 9 letters in all. The letter A occurs 4 times, the letter L occurs 2 times and the remaining three letters H, B, D each occurs once.
$\therefore$ Required number of permutations
$ =\frac{9 !}{4 ! 2 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 2 \times 1} $
$ =9 \times 8 \times 7 \times 3 \times 5 $ = 7560

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Question 212 Marks

If $\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}$, find x.

Answer

Given,$\frac{1}{8 !}+\frac{1}{9 \times 8 !}=\frac{x}{10 \times 9 \times 8 !}$
Therefore 1 + $\frac{1}{9}=\frac{x}{10 \times 9}$ or $\frac{10}{9}=\frac{x}{10 \times 9}$
Thus, x = 100

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Question 222 Marks

Evaluate $\frac{n !}{r !(n-r) !}$, when n = 5, r = 2

Answer

Now,to find $\frac{n !}{r !(n-r) !}$ (since n = 5, r = 2)
We have, $\frac{5 !}{2 !(5-2) !}=\frac{5 !}{2 ! \times 3 !}=\frac{5 \times 4}{2}=10$

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Question 232 Marks

Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.

Answer

Here, we have to find the number of different signals that can be generated by arranging at least 2 flags in order i.e., a signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. If a signal consists of 2 flags, then vacant places are 2 and 5 flags are available.
$\therefore$ Number of ways of filling first vacant place = 5
Number of ways of filling second vacant place = 4
Then, the total number of signals consisting by 2 flags
= 5 $\times$ 4 = 20
Now, if a signal consists of 3 flags, then vacant places are 3 and 5 flags are available.
$\therefore$ Number of ways of filling of first vacant place = 5
Number of ways of filling of second vacant place = 4
Number of ways of filling of third vacant place = 3
Then, the total number of signals consisting by 3 flags
= 5 $\times$ 4 $\times$ 3 = 60
Similarly, the total number of signals consisting of 4 flags
= 5 $\times$ 4 $\times$ 3 $\times$ 2 = 120
Total number of signals consisting of 5 flags
= 5 $\times$ 4 $\times$ 3 $\times$ 2 $\times$ 1 = 120
Now, different signals can be generated by arranging either 2 flags or 3 flags or 4 flags or 5 flags. So, by the fundamental principle of addition, we get
Total number of signals = 20 + 60 + 120 + 120 = 320

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Question 242 Marks

How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?

Answer

There will be as many ways as there are ways of filling 2 vacant places, blank in succession by the five given digits. Here, it is given that in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, therefore, the required number of two digits even numbers is 2 $\times$ 5, i.e., 10.

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Question 252 Marks

In how many ways can $5$ girls and 3 boys be seated in a row so that no two boys are together?

Answer

Let us assume that first seat the 5 girls.
This can be done in $5!$ ways. For each such arrangement,
the three boys can be seated only at the cross marked places.
$\times G \times G \times G \times G \times G \times.$
There are 6 cross marked places and the three boys can be seated in$\ ^6P_3$ ways.
Therefore, by multiplication principle,
the total number of ways $= 5! \times\ ^6P_3 = 5! \times \frac{6 !}{3 !}$
$= 4 \times 5 \times 2 \times 3 \times 4 \times 5 \times 6 = 14400$

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Question 262 Marks

Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?

Answer

There will be as many signals as there are ways of filling in 2 vacant places, blank in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by any one of the 4 flags; we have, the lower vacant place can be filled in 3 different ways by any one of the remaining 3 different flags. Therefore, by the multiplication principle, the required number of signals = 4 $\times$ 3 = 12.

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Question 272 Marks

A committee of $3$ persons is to be constituted from a group of $2$ men and $3$ women. In how many ways can this be done? How many of these committees would consist of $1$ man and $2$ women$?$

Answer

Here, order does not matter. Thus, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken $3$ at a time. Therefore, the required number of ways $ = ^{5} C_{3}=\frac{5 !}{3 ! 2 !}=\frac{4 \times 5}{2}=10$.
Now, $1$ man can be selected from $2$ men in $^2C_1$ ways and 2 women can be selected from $3$ women in $^3C_2$ ways. Thus, the required number of committees $=\ ^{2} \mathrm{C}_{1} \times^{3} \mathrm{C}_{2}=\frac{2 !}{1 ! !} \times \frac{3 !}{2 ! 1 !}=6$

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Question 282 Marks

If $^nC_9 =\ ^nC_8,$ find $^nC_{17}$

Answer

We have, $^nC_9 =\ ^nC_8$
i.e., $\frac{n !}{9 !(n-9) !}=\frac{n !}{(n-8) ! 8 !}$
or $\frac{1}{9}=\frac{1}{n-8}$ or $n - 8 = 9$ or $n = 17$
Thus, $^nC_{17} =\ ^{17}C_{17} = 1$

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Question 292 Marks

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?

Answer

We know that the total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Thus, the number of arrangements $\frac{9 !}{4 ! \ 3 ! \ 2 !}=1260$

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Question 302 Marks

Find the number of different $8-$letter arrangement that can be made from the letters of the word DAUGHTER so that all vowels occur together.

Answer

We know that There are $8$ different letters in the word DAUGHTER, in which there are $3$ vowels, namely, $A, U$ and $E.$
Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE).
This single object together with $5$ remaining letters (objects) will be counted as $6$ objects.
Then we count permutations of these $6$ objects taken all at a time.
Therefore, this number would be $^6P_6 = 6!$. Corresponding to each of these permutations, we shall have $3!$ permutations of the three vowels $A, U, E$ taken all at a time.
Therefore, by the multiplication principle, the required number of permutations is $= 6 !\times 3! = 4320$

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Question 312 Marks

How many numbers lying between $100$ and $1000$ can be formed with the digits $0, 1, 2, 3, 4, 5,$ if the repetition of the digits is not allowed?

Answer

We know that,every number between $100$ and $1000$ is a $3-$digit number.
We, first, have to count the permutations of $6$ digits taken 3 at a time.
This number would be $^6P_3.$
But, these permutations will include those also where $0$ is at the $100’s$ place.
For example, $092, 042, . . .,$ etc are such numbers which are actually $2-$digit numbers and therefore the number of such numbers has to be subtracted from $^6P_3 $ to get the required number.
To get the number of such numbers, we fix $0$ at the $100’s$ place and rearrange the remaining $5$ digits taking $2$ at a time.
This number is $^5P_2 .$
Therefore, The required number will be =$^{6} \mathrm{P}_{3}-^{5} \mathrm{P}_{2}=\frac{6 !}{3 !}-\frac{5 !}{3 !}$
$= 4 \times 5 \times 6 – 4 \times 5 = 100$

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Question 322 Marks

How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?

Answer

Here order matters for example 1234 and 1324 are two different numbers. Thus, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Thus, the required 4 digit numbers = $^{9} \mathrm{P}_{4}=\frac{9 !}{(9-4) !}=\frac{9 !}{5 !}=9 \times 8 \times 7 \times 6=3024$

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Question 332 Marks

Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.

Answer

We have, there are as many words as there are ways of filling in 4 vacant places, blank by the 4 letters, but it is given that the repetition is not allowed. The first place can be filled in 4 different ways by any one of the 4 letters R, O, S, E. Following which, the second place can be filled in by any one of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Therefore, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 $\times$ 3 $\times$ 2 $\times$ 1 = 24.
Therefore, the required number of words is 24.

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