SEBA Class 10 Advanced Maths Chapter 6.3 Solutions: Restricted Permutations

Looking for reliable and step-by-step SEBA Class 10 Advanced Maths Chapter 6.3 Solutions? Here, you will get SEBA Class 10 Advanced Maths Chapter 6.3 Solutions for HSLC exams. In this article, we solve all the textbook exercises of Chapter 6.3: Restricted Permutations, from the SEBA Class 10 New Advanced Mathematics book. We provide all solutions of Chapter 6.3 in a very simple way to help you score high marks in your upcoming HSLC board examinations.

See also
SEBA Class 10 Advanced Maths Chapter 6.1
SEBA Class 10 Advanced Maths Chapter 6.2
SEBA Class 10 Advanced Maths Chapter 6.4

Q1: How many number of 5 digits can be formed without repetition if :

(i) 2, 3 and 5 always occur in each number?

(ii) 0 never occurs.

Soln :

(i) When 2, 3 and 5 always occur in each number:

The total available decimal digits are: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

$\therefore$ Total number of digits, $n = 10$.

We need to form 5-digit numbers, so $r = 5$.

Here, $k = 3$ ( always occur)

We know that the number of permutations where $k$ particular objects always occur is ${}^{n-k}P_{r-k} \times {}^rP_k$.

$\therefore$ Total number of arrangements:

$= {}^{10-3}P_{5-3} \times {}^5P_3$

$= {}^7P_2 \times {}^5P_3$

$= \frac{\lfloor 7}{\lfloor 7-2} \times \frac{\lfloor 5}{\lfloor 5-3}$

$= \frac{\lfloor 7}{\lfloor 5} \times \frac{\lfloor 5}{\lfloor 2}$

$= \frac{7 \times 6 \times \lfloor 5}{\lfloor 5} \times \frac{5 \times 4 \times 3 \times \lfloor 2}{\lfloor 2}$

$= (7 \times 6) \times (5 \times 4 \times 3)$

$= 42 \times 60$

$= 2520$

$= 42 \times 60$

$= 2520$

Now, among these permutations, some numbers have $0$ in the first position (ten-thousands place). Such numbers are 4-digit numbers.

[When $0$ is fixed in the first place:

Remaining places to fill $= 5 – 1 = 4$.

Remaining digits available $= 10 – 1 = 9$.

Since the 3 digits ($2, 3, 5$) must still be present:]

$\therefore$ Number of arrangements having 0 at the first place

$= {}^{9-3}P_{4-3} \times {}^4P_3$

$= {}^6P_1 \times {}^4P_3$

$= 6 \times (4 \times 3 \times 2)$

$= 6 \times 24$

$= 144$

$\therefore$ Required number of 5-digit numbers formed

$= 2520 – 144$

$= 2376$

(ii) When 0 never occurs:

Since $0$ never occurs, $0$ is completely excluded.

$\therefore$ Remaining digits available $= 10 – 1 = 9$ (i.e., $1, 2, 3, 4, 5, 6, 7, 8, 9$).

$\therefore$ Required number of 5-digit numbers

$= {}^9P_5$

$= 9 \times 8 \times 7 \times 6 \times 5$

$= 15120$

Q2: In how many ways 10 different books taken 4 at a time can be arranged in an almirah such that a particular book-
(i) is always taken?
(ii) is never taken?

Soln :

Total number of books, $n = 10$.

Number of books taken at a time, $r = 4$.

(i) When a particular book is always taken:

Here, $k = 1$ , since a particular book is always included.

$\therefore$ Required number of ways;

${}^{n-k}P_{r-k} \times {}^rP_k$:

$= {}^{10-1}P_{4-1} \times {}^4P_1$

$= {}^9P_3 \times 4$

$= (9 \times 8 \times 7) \times 4$

$= 504 \times 4$

$= 2016$

(ii) When a particular book is never taken:

Here, $k = 1$, since a particular book is never included.

$\therefore$ Required number of ways

${}^{n-k}P_r$

$= {}^{10-1}P_4$

$= {}^9P_4$

$= 9 \times 8 \times 7 \times 6$

$= 3024$

Q3: How many numbers can be formed by the digits 1, 2, 3, 4 and 5 if the numbers are less than 1000 and repetition is allowed for each digit.

Soln :

1-digit numbers that can be formed $=5$

2-digit numbers that can be formed $= 5 \times 5 = 25$.

3-digit numbers that can be formed $= 5 \times 5 \times 5 = 125$.

$\therefore$ Total required numbers

$= 5 + 25 + 125$

$= 155$

Q4: From a school 5 boys and 4 girls has been selected for a competition. How many ways photographs can be taken placing them in a row such that no two girls remain together.

Soln :

$\_ \text{ B } \_ \text{ B } \_ \text{ B } \_ \text{ B } \_ \text{ B } \_$

The 5 boys can be arranged in a row in $5!=120$ ways.

Since no two girls sit together. So, 6 gaps are formed, including the two ends.

These 6 can be filled by the 4 girls in ${}^6P_4$ ways.

$= {}^6P_4$

$= 6 \times 5 \times 4 \times 3$

$= 360\text{ ways}$.

$\therefore$ Total required number of ways:

$= 120 \times 360$

$= 43200$

Q5: How many ways the letters of the word ORIENTAL can be arranged if

(i) no two vowels come together?

(ii) consonants and vowels are placed one by one?

Soln :

The word $\text{ORIENTAL}$ contains 8 letters.

Vowels are: $\text{O, I, E, A}$ (4 vowels).

Consonants are: $\text{R, N, T, L}$ (4 consonants).

(i) When no two vowels come together:

The 4 consonants can be arranged in $4! = 24\text{ ways}$.

The 4 consonants create 5 gaps:

$\_ \text{ C } \_ \text{ C } \_ \text{ C } \_ \text{ C } \_$

The 4 vowels can be placed in any 4 of these 5 gaps in ${}^5P_4 \text{ ways}$.

${}^5P_4 = 5 \times 4 \times 3 \times 2 = 120\text{ ways}$.

$\therefore$ Total required arrangements:

$= 24 \times {}^5P_4$

$= 24 \times 5 \times 4 \times 3 \times 2$

$= 24 \times 120$

$= 2880$

(ii) When consonants and vowels are placed one by one (alternately):

Since the number of vowels and consonants are equal (4 each), the arrangements can be formed in 2 patterns:

• $\text{C V C V C V C V}$
• $\text{V C V C V C V C}$

The 4 consonants can be arranged in $4! = 24\text{ ways}$.

Similarly, 4 vowels can be arranged in $4! = 24\text{ ways}$.

Also, the word can either start with a consonant OR start with a vowel.

∴ The total arrangements:

$= 1152$

Q6: How many words can be formed by the letters of the words POLYHEDRA if

(i) the vowels do not come together?

(ii) no two vowels come together?

Soln :

The word $\text{POLYHEDRA}$ has 9 letters.

Vowels: $\text{O, E, A}$ (3 vowels).

Consonants: $\text{P, L, Y, H, D, R}$ (6 consonants).

(i) When the vowels do not come together:

Total number of words formed by taking all 9 letters without restriction

$= 9! = 362880$.

When all 3 vowels come together:

Let us consider the 3 vowels $(\text{O, E, A})$ as a single unit.

Then the number of letters becomes $6 + 1 = 7$.

These 7 letters can be arranged in $7! = 5040\text{ ways}$.

Also, the 3 vowels can be arranged among themselves in $3! = 6\text{ ways}$.

$\therefore$ Number of words where vowels always come together

$= 7! \times 3!$

$= 5040 \times 6$

$= 30240$

$\therefore$ Number of words where vowels do not come together

$= 9! – (7! \times 3!)$

$= 362880 – 30240$

$= 332640$

(ii) When no two vowels come together:

The 6 consonants can be arranged in $6! = 720\text{ ways}$.

The 6 consonants create 7 gaps:

$\_ \text{ C } \_ \text{ C } \_ \text{ C } \_ \text{ C } \_ \text{ C } \_ \text{ C } \_$

The 3 vowels can be placed in any 3 of these 7 gaps in ${}^7P_3\text{ ways}$

$\therefore$ Total required arrangements

$= 6! \times {}^7P_3$

$= 720 \times 7 \times 6 \times 5$

$= 720 \times 210$

$= 151200$

Q7: How many words can be formed by the letters of the word INDEPENDENCE. In how many of them the vowels do not occur together?

Soln :

The word $ \text{INDEPENDENCE} $ has total 12 letters. There are 3 ‘$\text{N}$’, 3 ‘$\text{D}$’ and 4’$\text{E}$’ and the remaining letters $\text{I, P, C}$ occur 1 time each

Total number of words that can be formed:

$= \frac{\lfloor 12}{\lfloor 3 \cdot \lfloor 2 \cdot \lfloor 4}$

$= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times \lfloor 4}{(3 \times 2 \times 1) \times (2 \times 1) \times \lfloor 4}$

$= \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{6 \times 2}$

$= \frac{39916800}{12}$

$= 1663200$

Part 2: When the vowels do not occur together

Let us consider all 5 vowels $\text{E, E, E, E, I}$ as a single unit.

Then the total number of units to be arranged $= 7\text{ consonants} + 1\text{ vowel unit} = 8\text{ units}$.

These 8 units (where $\text{N}$ appears 3 times and $\text{D}$ appears 2 times) can be arranged in:

$= \frac{\lfloor 8}{\lfloor 3 \cdot \lfloor 2}$

$= \frac{8 \times 7 \times 6 \times 5 \times 4 \times \lfloor 3}{\lfloor 3 \times (2 \times 1)}$

$= \frac{8 \times 7 \times 6 \times 5 \times 4}{2}$

$= \frac{6720}{2}$

$= 3360\text{ ways}$

Also, the 5 vowels can be arranged among themselves in:

$= \frac{\lfloor 5}{\lfloor 4}$

$= \frac{5 \times \lfloor 4}{\lfloor 4}$

$= 5\text{ ways}$

$\therefore$ Number of words where vowels always occur together:

$= 3360 \times 5$

$= 16800$

$\therefore$ Number of words in which vowels do not occur together:

$= \text{Total words} – \text{Words where vowels occur together}$

$= 1663200 – 16800$

$= 1646400$

Q8: How many words can be formed by the letters of the word PERMUTATIONS if-
(i) the words begin with P and end with S?
(ii) the vowels always occur together?
(iii) there be always 4 letters between P and S?

Soln :

There are 12 letters in the word $\text{PERMUTATIONS}$, of which $\text{T}$ occurs 2 times and other letters occur once.

(i) When the words begin with P and end with S:

Fixing $\text{P}$ in the 1st position and $\text{S}$ in the 12th position, we have $12 – 2 = 10$ places left to fill with the remaining 10 letters (which include two T’s).

$\therefore$ Required number of words: $= \frac{\lfloor 10}{\lfloor 2}$

$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times \lfloor 2}{\lfloor 2}$

$= 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3$

$= 1814400$

(ii) When the vowels always occur together:

Let us consider the 5 vowels $\text{A, E, I, O, U}$ as a single unit. Then the total units to arrange $= 7\text{ consonants} + 1\text{ vowel unit} = 8\text{ units}$.

These 8 units (having two T’s) can be arranged in:

$= \frac{\lfloor 8}{\lfloor 2}$

$= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times \lfloor 2}{\lfloor 2}$

$= 8 \times 7 \times 6 \times 5 \times 4 \times 3$

$= 20160\text{ ways}$

Also, the 5 distinct vowels can be arranged among themselves in:

$= \lfloor 5$

$= 5 \times 4 \times 3 \times 2 \times 1$

$= 120\text{ ways}$

$\therefore$ Required number of words where the vowels always occur together:

$= 20160 \times 120$

$= 2419200$

(iii) When there be always 4 letters between P and S:

The positions of $\text{P}$ and $\text{S}$ can be: $(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$

$\therefore$ To fill P and S, we have $= 7$ choices

Also $\text{P}$ and $\text{S}$ can arrange themselves in $\lfloor 2$ ways.

∴ Number of ways to place $\text{P}$ and $\text{S} = 7 \times \lfloor 2 = 7 \times 2 = 14\text{ ways}$.

The remaining 10 places can be filled by the remaining 10 letters (having two T’s) in:

$= \frac{\lfloor 10}{\lfloor 2}$ ways

$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times \lfloor 2}{\lfloor 2}$

$= 1814400\text{ ways}$

$\therefore$ Total required number of words that can be formed :

$= 14 \times 1814400$

$= 25401600$

Q9: How many ways the letters of the word EQUATION can be arranged if the vowels and consonants always occur together?

Soln :

The word $\text{EQUATION}$ has 8 letters.

Vowels: $\text{E, U, A, I, O}$ (5 vowels).

Consonants: $\text{Q, T, N}$ (3 consonants).

Let us assume the group of vowels as 1 unit and the group of consonants as 1 unit.

These 2 units can be arranged in: $= \lfloor 2\text{ ways}$ 

The 5 vowels can be arranged among themselves in: $= \lfloor 5\text{ ways}$ 

The 3 consonants can be arranged among themselves in: $= \lfloor 3\text{ ways}$ 

$\therefore$ Required number of arrangements when the vowels and consonants always occur together:

$= \lfloor 2 \times \lfloor 5 \times \lfloor 3$

$= 2 \times 120 \times 6$

$= 1440$

Q10: Show that the total permutations of $n$ different objects taking $r$ at a time is $(r-k+1)\times{}^{n-k}P_{r-k}$ if $k$ particular object always remain together with certain order.

Proof :

If, initially, the $k$ particular objects which occur together are kept aside.

Then the remaining $n – k$ objects can fill the remaining places $r – k$ in: $ {}^{n-k}P_{r-k}\text{ ways}$

Now, the $k$ particular objects which occur together are considered as a single object.

This single object can be put with the $ {}^{n-k}P_{r-k}$ permutation in : $ (r – k) + 1 = (r – k + 1)\text{ ways}$.

As the $k$ particular objects must always remain together in a fixed order, they can be arranged among themselves in only: $1\text{ way}$.

$\therefore$ The total number of permutations is:

$= {}^{n-k}P_{r-k} \times (r – k + 1) \times 1$

$= (r – k + 1) \times {}^{n-k}P_{r-k}$

Hence proved.