3 Marks Question

Question 13 Marks

It is required, to seat 5 men, and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Answer

Here we have 5 men and 4 women. It is given that women occupy the even places.

1	2	3	4	5	6	7	8	9
M	W	M	W	M	W	M	W	M

Here we have four even places. So we can arrange four womens in 4! ways and 5 men in 5! ways in remaining five places.
$\therefore $ Required number of ways $ = 4! \times 5!$
$ = 24 \times 120 = 2880$

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Question 23 Marks

Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Answer

Here we have total cards = 52
We have to select 5 cards containing 1 king and 4 other cards i.e. we have to select 1 king out of 4 kings and 4 other cards out of 48 other cards.
$\therefore $ Number of selections ${ = ^4}{C_1}{ \times ^{48}}{C_4}$
$\frac{{4!}}{{3!1!}} \times \frac{{48!}}{{4!44!}}$
$ = \frac{{4 \times 3!}}{{3! \times 1}} \times \frac{{48 \times 47 \times 46 \times 45 \times 44!}}{{4 \times 3 \times 2 \times 1 \times 44!}}$
$ = 4 \times 2 \times 47 \times 46 \times 45 = 778320$

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Question 33 Marks

In an examination a question paper consist of 12 questions divided into two parts i.e. part I and part II containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Answer

Here number of questions in part I are 5 and number of questions in part II are 7. We have to select 8 questions at least 3 questions from each section. So we have required selections are 3 from part I and 5 from part II or 4 from part I and 4 from part II or 5 from part I and 3 from part II.
$\therefore $ Number of ways of selection ${ = ^5}{C_3}{ \times ^7}{C_5}{ + ^5}{C_4}{ \times ^7}{C_4}{ + ^5}{C_5}{ \times ^7}{C_3}$
$ = \frac{{5!}}{{3!2!}} \times \frac{{7!}}{{5!2!}} + \frac{{5!}}{{4!1!}} \times \frac{{7!}}{{4!3!}}$$ + \frac{{5!}}{{5!0!}} \times \frac{{7!}}{{5!0!}}$
$ = \frac{{5 \times 4 \times 3!}}{{3! \times 2 \times 1}} \times \frac{{7 \times 6 \times 5!}}{{5! \times 2 \times }} + \frac{{5 \times 4!}}{{4! \times 1}}$$ \times \frac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2 \times 1}} + 1 \times \frac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}}$
$ = 10 \times 21 + 5 \times 35 + 1 \times 35$
= 210 + 175 + 35 = 420

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Question 43 Marks

The English alphabet has 5 vowels and 21 consonants. How many words with 2 vowels and 2 different consonants can be formed from the alphabet?

Answer

Here we have 5 vowels and 21 consonants. We have to select 2 vowels out of 5 vowels and 2 consonants out of 21 consonants and these 4 letters can be arranged in 4! ways.
$\therefore $ Required number of words ${ = ^5}{C_2}{ \times ^{21}}{C_2} \times 4!$
$ = \frac{{5!}}{{2!3!}} \times \frac{{21!}}{{2!19!}} \times 4!$
$ = \frac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} \times \frac{{21 \times 20 \times 19!}}{{2 \times 1 \times 19!}} \times 4 \times 3 \times 2 \times 1$
$ = 10 \times 210 \times 24 = 50400$

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Question 53 Marks

How many $6-$digit numbers can be formed from the digits $0, 1, 3, 5, 7$ and $9$ which are divisible by $10$ and, no digit is repeated?

Answer

A number divisible by $10$ have unit place digit $0.$ So digit $0$ is fixed at unit place and the remaining $5$ places can be filled with remaining five digits in $ ^5P_5 $ ways.
$\therefore $ Required numbers $ ^5P_5= 120$

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Question 63 Marks

If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Answer

In this problem, we have to find the number of words starting with E. Here in EXAMINATION we have two I's and two N's and all other letters are different.
$\therefore $ Number of ways of arrangement $ = \frac{{10!}}{{2!2!}}$
$ = \frac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}}{{2 \times 1 \times 2!}}$= 907200

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Question 73 Marks

How many word with or without meaning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Answer

There are 8 letters in the word EQUATION including 5 vowels and 3 consonants. Now 5 vowels can be arranged in 5! Ways and 3 consonants can be arranged in 3! Ways. Also the two groups of vowels and consonants can be arranged in 2! Ways.
$\therefore $ Total number of permutation $ = 5! \times 3! \times 2!$
$ = 120 \times 6 \times 2 = 1440$

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Question 83 Marks

From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them, will join or none of them will join. In how many ways can the excursion party be chosen?

Answer

Here total students are 25 from which 10 are chosen for an excursion party. If the 3 students join the party then 7 students out of remaining 22 students or 3 students not join the party them 10 students out of remaining 22 students.
$\therefore $ Number of ways of selection ${ = ^3}{C_3}{ \times ^{22}}{C_7}{ + ^3}{C_0}{ \times ^{22}}{C_{10}}$
$ = 1 \times \frac{{22!}}{{7!15!}} + 1 \times \frac{{22!}}{{10!12!}}$
$ = \frac{{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 15!}}$$ + \frac{{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}}{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}}$
= 170544 + 646646 = 817190

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Question 93 Marks

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER.

Answer

There are 8 letters in the word DAUGHTER including 3 vowels and 5 consonants. We have to select 2 vowels out of 3 vowels and 3 consonants of 5 consonants.
$\therefore $ Number of ways of selection ${ = ^3}{C_2}{ \times ^5}{C_3} = 3 \times 10 = 30$
Now each word contains 5 letters which can be arranged among themselves in 5! Ways.
So total number of words $ = 5! \times 30 = 120 \times 30 = 3600$

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Question 103 Marks

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer

There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour
$\therefore $ Number of ways of selection ${ = ^6}{C_{_3}}{ \times ^5}{C_{_3}}{ \times ^5}{C_{_3}}$
$= \frac{{6!}}{{3!3!}} \times \frac{{5!}}{{3!2!}} \times \frac{{5!}}{{3!2!}} = $$20 \times 10 \times 10 = 2000$

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Question 113 Marks

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

Answer

Total letters of the word MISSISSIPPI = 11
Here M =1, I = 4, S = 4 and P = 2
$\therefore $ Number of permutations = $\frac{{11!}}{{4!4!2!}}$
= $\frac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}}$ = 34650
When the four I's come together then it becomes one letter so total number of letters in the word when all I's come together = 8
$\therefore $ Number of Permutations = $\frac{{8!}}{{4!2!}} = \frac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 2 \times 1}}$ = 840
Number of permutations when four I's do not come together = 34650 - 840 = 33810

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Question 123 Marks

How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?

Answer

We know that 1000000 is a 7-digit number and the number of digits to be used is also
Thus, the numbers to be counted will be 7 -digit only. Also, the numbers have to be greater than 1000000, therefore they can begin either with 1, 2 or 4

The number of numbers beginning with $1=\frac{6 !}{3 ! 2 !}=\frac{4 \times 5 \times 6}{2}=60$, as when 1 is
fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4, in which there are 3, 2 s and 2, 4 s
Total numbers begining with 2
= $\frac{6 !}{2 ! 2 !}=\frac{3 \times 4 \times 5 \times 6}{2}=180$
and total numbers begining with $4=\frac{6 !}{3 !}=4 \times 5 \times 6=120$
Thus,the required number of numbers = 60 + 180 + 120 = 360

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Question 133 Marks

Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the $50^{th}$ word?

Answer

We have, There are $5$ letters in the word AGAIN, in which A appears $2$ times. Thus,
the required number of words = $\frac{5 !}{2 !} = 60$
To get the number of words starting with $A,$ we fix the letter $A$ at the extreme left position, we then rearrange the remaining $4$ letters taken all at a time. There will be as many arrangements of these $4$ letters taken $4$ at a time as there are permutations of $4$ different things taken $4$ at a time. Therefore, the number of words starting with $A = 4! = 24.$
Then, starting with $G,$ the number of words = $\frac{4 !}{2 !} = 12$ as after placing $G$
at the extreme left position, we are left with the letters $A, A, I$ and $N.$ Similarly, there are $12$ words starting with the next letter I. Therefore,total number of words so far obtained $= 24 + 12 + 12 = 48$
Therefore, the $49^{th}$ word is NAAGI and he $50^{th}$ word is NAAIG.

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Question 143 Marks

How many words, with or without meaning, each of $3$ vowels and $2$ consonants can be formed from the letters of the word INVOLUTE$?$

Answer

We have, in the word INVOLUTE, there are $4$ vowels, namely, $I, O, E, U$ and $4$ consonants, namely, $N, V, L$ and $T.$
The number of ways of selecting $3$ vowels out of $4 =\ ^4C_3 = 4.$
The number of ways of selecting $2$ consonants out of $4 =\ ^4C_2 = 6$. Thus, the number of combinations of $3$ vowels and $2$ consonants is $4 \times 6 = 24.$
Now, we each of these $24$ combinations has 5 letters which can be arranged among themselves in $5 !$ ways. Thus, the required number of different words is $24 \times 5 ! = 2880$

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Question 153 Marks

Find: $r,$ if $5\ ^4P_r = 6\ ^5P_{r - 1}$

Answer

We have, $5.\ ^4P_r = 6 . (^5P_{r - 1})$
$\Rightarrow 5 \cdot \frac{4 !}{(4-r) !}=6 \times \frac{5 !}{[5-(r-1)] !} $
$ \Rightarrow \frac{5 \cdot 4 !}{(4-r) !}=\frac{6 \times 5 \times 4 !}{(6-r) !} $
$ \Rightarrow \quad \frac{1}{(4-r) !}=\frac{6}{(6-r)(5-r)(4-r) !} $
$\Rightarrow (6 - r) (5 - r) = 6$
$\Rightarrow 30 - 11r + r^2 = 6$
$\Rightarrow r^2 - 11r + 24 = 0$
$\Rightarrow (r - 3) (r - 8) = 0$
$\Rightarrow r = 3, 8$
But $r \neq 8,$ because in $^4P_r,$ r cannot be greater than $4.$
Hence, $r = 3$

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