SEBA Class 10 Advanced Maths Chapter 6.2 Solutions: 

Here, you will get simple and reliable SEBA Class 10 Advanced Maths Chapter 6.2 Solutions for Class 10 students from the Assam SEBA board. These solutions are from Chapter 6.2 from the SEBA Class 10 New Advanced Mathematics book. We provide all solutions of Chapter 6.2 in a very simple way for the upcoming HSLC Matric examination.

For better understanding.
See also
SEBA Class 10 Advanced Maths Chapter 6.1

1. Find the value of :

(i) $^{15}P_{3}$

Solution:

$^{15}P_{3} $

$= \frac{\lfloor 15}{\lfloor 15-3} $

$= \frac{\lfloor 15}{\lfloor 12} $

$= \frac{15 \times 14 \times 13 \times \lfloor 12}{\lfloor 12} $

$= 15 \times 14 \times 13 $

$= 2730$

(ii) $^{11}P_{5}$

Solution:

$^{11}P_{5} $
$= \frac{\lfloor 11}{\lfloor 11-5} $
$= \frac{\lfloor 11}{\lfloor 6} $
$= \frac{11 \times 10 \times 9 \times 8 \times 7 \times \lfloor 6}{\lfloor 6} $
$= 11 \times 10 \times 9 \times 8 \times 7 $
$= 55440$

(iii) $^{8}P_{0}$

Solution:

$^{8}P_{0} $
$= \frac{\lfloor 8}{\lfloor 8-0} $
$= \frac{\lfloor 8}{\lfloor 8} $
$= 1$

(iv) $^{43}P_{38}$

Solution:

$^{43}P_{38} $
$= \frac{\lfloor 43}{\lfloor 43-38} $
$= \frac{\lfloor 43}{\lfloor 5}$

2. Show that-

(i) $^{n}P_{r} + r \cdot ^{n}P_{r-1} = {^{n+1}P_{r}}$

Proof:

L.H.S. $= {^{n}P_{r}} + r \cdot {^{n}P_{r-1}}$

$= \frac{\lfloor n}{\lfloor n-r} + r \cdot \frac{\lfloor n}{\lfloor n-(r-1)}$

$= \frac{\lfloor n}{\lfloor n-r} + \frac{r \lfloor n}{\lfloor n-r+1}$

$= \frac{\lfloor n}{\lfloor n-r} + \frac{r \lfloor n}{(n-r+1)\lfloor n-r}$

$= \frac{\lfloor n}{\lfloor n-r} \left(1 + \frac{r}{n-r+1}\right)$

$= \frac{\lfloor n}{\lfloor n-r} \left(\frac{n-r+1+r}{n-r+1}\right)$

$= \frac{\lfloor n}{\lfloor n-r} \cdot \frac{n+1}{n-r+1}$

$= \frac{(n+1)\lfloor n}{(n-r+1)\lfloor n-r}$

$= \frac{\lfloor n+1}{\lfloor (n+1)-r}$

$= {^{n+1}P_{r}} =$ R.H.S.

Hence proved.

(ii) $n(n-1) \cdot {^{n-2}P_{r-2}} = {^{n}P_{r}}$

Proof:

L.H.S. $= n(n-1) \cdot {^{n-2}P_{r-2}}$

$= n(n-1) \cdot \frac{\lfloor n-2}{\lfloor (n-2)-(r-2)}$

$= n(n-1) \cdot \frac{\lfloor n-2}{\lfloor n-2-r+2}$

$= \frac{n(n-1)\lfloor n-2}{\lfloor n-r}$

$= \frac{\lfloor n}{\lfloor n-r}$

$= {^{n}P_{r}} =$ R.H.S.

Hence proved.

(iii) $^{n}P_{r} = (n-r+1) \cdot {^{n}P_{r-1}}$

Proof:

R.H.S. $= (n-r+1) \cdot {^{n}P_{r-1}}$

$= (n-r+1) \cdot \frac{\lfloor n}{\lfloor n-(r-1)}$

$= (n-r+1) \cdot \frac{\lfloor n}{\lfloor n-r+1}$

$= (n-r+1) \cdot \frac{\lfloor n}{(n-r+1)\lfloor n-r}$

$= \frac{\lfloor n}{\lfloor n-r}$

$= {^{n}P_{r}} =$ L.H.S.

Hence proved.

(iv) $1 \cdot {^{1}P_{1}} + 2 \cdot {^{2}P_{2}} + 3 \cdot {^{3}P_{3}} + \dots + n \cdot {^{n}P_{n}} = {^{n+1}P_{n+1}} – 1$

Proof:

We know that $^{r}P_{r} = \lfloor r$

$\therefore r \cdot {^{r}P_{r}} = r \cdot \lfloor r $
$ = (r+1-1)\lfloor r $
$ = (r+1)\lfloor r – \lfloor r $
$ = \lfloor r+1 – \lfloor r$

Putting $r = 1, 2, 3, \dots, n$ successively:

$1 \cdot {^{1}P_{1}} = \lfloor 2 – \lfloor 1$

$2 \cdot {^{2}P_{2}} = \lfloor 3 – \lfloor 2$

$3 \cdot {^{3}P_{3}} = \lfloor 4 – \lfloor 3$

$\dots \dots \dots$

$n \cdot {^{n}P_{n}} = \lfloor n+1 – \lfloor n$

Adding all these, we get:

L.H.S. $= 1 \cdot {^{1}P_{1}} + 2 \cdot {^{2}P_{2}} + 3 \cdot {^{3}P_{3}} + \dots + n \cdot {^{n}P_{n}}$

$= (\lfloor 2 – \lfloor 1) + (\lfloor 3 – \lfloor 2) + (\lfloor 4 – \lfloor 3) + \dots + (\lfloor n+1 – \lfloor n)$

$= \lfloor n+1 – \lfloor 1$

$= {^{n+1}P_{n+1}} – 1 =$ R.H.S.

Hence proved.

3. How many numbers between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5 without repetition of any digit?

Solution:

Clearly, the numbers lying between 100 and 1000 are all 3-digit numbers.

Total number of given digits $= 5$ (which are $1, 2, 3, 4, 5$).

We need to form 3-digit numbers without repetition.

$\therefore$ Required numbers $= {^{5}P_{3}} $
$ = \frac{\lfloor 5}{\lfloor 5-3} $
$ = \frac{\lfloor 5}{\lfloor 2} $
$ = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} $
$ = 60$

4. How many numbers of 4 digits can be formed with the odd digits without repeating any digit?

Solution:

The odd digits are $1, 3, 5, 7, 9$.

Total number of odd digits $= 5$.

We need to form 4-digit numbers without repetition.

$\therefore$ Required numbers $= {^{5}P_{4}} $
$ = \frac{\lfloor 5}{\lfloor 5-4} $
$ = \frac{\lfloor 5}{\lfloor 1} $
$ = 5 \times 4 \times 3 \times 2 \times 1 $
$ = 120$

5. How many odd numbers of 5 digits can be formed with the digits 0, 2, 3, 4, 5 without repeating any digit?

Solution:

Total number of given digits $= 5$

For a 5-digit number to be odd, the unit’s place must be filled by an odd digit ($3$ and $5$).

So, the unit’s place can be filled in $2$ ways.

Since repetition is not allowed,

The first place (ten thousandth) cannot be $0$.

So, the first place can be filled in $3$ ways.

The remaining $3$ places (thousandth, hundredth, and tenth places) can be filled by the remaining $3$ digits in $^{3}P_{3} = \lfloor 3 $ ways.

$\therefore$ Required number of odd numbers $= 3 \times \lfloor 3 \times 2 $
$= 3 \times 3 \times 2\times 2 $
$ = 36$

6. Without repeating any digit, how many 3 digit number can be formed with the odd digits.

Solution:

The odd digits are $1, 3, 5, 7, 9$ (total 5 digits).

Since repetition of digits is not allowed, the number of 3-digit numbers is the permutation of 5 digits taken 3 at a time.

$\therefore$ Required numbers $= {^{5}P_{3}} $
$ = 5 \times 4 \times 3 $
$ = 60$

7. How many different words with or without meaning can be formed by the letters of the word MONDAY begining with A and not ending with Y?

Solution:

There are $6$ letters in the word “MONDAY”.

There are 6 positions to fill: 1st, 2nd, 3rd, 4th, 5th, 6th.

Since the word must begin with A, the 1st position is fixed with A in 1 way.

The last position (6th) cannot be filled with Y, so it can be filled by any of the remaining 4 letters (M, O, N, D) in 4 ways.

Now, the remaining 4 positions can be filled by the remaining 4 letters in $^{4}P_{4} = \lfloor 4$ ways.

$\therefore$ Required number of words $= 1 \times \lfloor 4 \times 4 $
$ = 4 \times 4 \times 3 \times 2 \times 1$
$ = 4 \times 24 $
$ = 96$

8. In how many ways 5 boys and 6 girls can be seated in a long bench so that each boy will sit between two girls?

Solution:

6 girls can be seated in 6 places in $^{6}P_{6} = \lfloor 6$ ways.

When 6 girls are seated, they create 5 gaps between them:

$\text{G}_1 \text{ \_ } \text{G}_2 \text{ \_ } \text{G}_3 \text{ \_ } \text{G}_4 \text{ \_ } \text{G}_5 \text{ \_ } \text{G}_6$

5 boys can be seated in these 5 gaps in $^{5}P_{5} = \lfloor 5$ ways.

$\therefore$ Required number of ways $= \lfloor 6 \times \lfloor 5 $
$ = 720 \times 120 $
$ = 86400$

9. How many words can be formed with the letters of the word PENCIL begining with C?

Solution:

There are 6 letters in the word PENCIL.

Since each word must begin with C, the first place is fixed with letter C in 1 way.

The remaining 5 places can be filled by the remaining 5 letters (P, E, N, I, L) in $^{5}P_{5} = \lfloor 5$ ways.

$\therefore$ Required number of words $= 1 \times \lfloor 5 = 120$

10. How many ways the letters of the word RUBBER can be arranged?

Solution:

The word RUBBER contains 6 letters, of which there are 2 ‘R’s, 2 ‘B’s.

$\therefore$ Required number of arrangements $= \frac{\lfloor 6}{\lfloor 2 \times \lfloor 2} = \frac{720}{2 \times 2} $
$ = \frac{720}{4} $
$ = 180$

11. How many words not ending with W can be formed with the letters of the word FLOWER.

Solution:

There are 6 letters in the word ‘FLOWER’.

Total number of words that can be formed by all 6 letters $= {^{6}P_{6}} = \lfloor 6 = 720$.

Number of words ending with W $= {^{5}P_{5}} = \lfloor 5 = 120$ (Since W is fixed at unit place, the remaining 5 places can be filled by the remaining 5 letters.)

$\therefore$ Number of words not ending with W $= \text{Total words} – \text{Words ending with W} $
$= 720 – 120 $
$= 600$

12. How many ways 5 towels can be put into 4 boxes?

Solution:

The first towel can be put into any of the 4 boxes in 4 ways.

The second towel can be put into any of the 4 boxes in 4 ways.

Similarly, the third, fourth, and fifth towels can each be put into any of the 4 boxes in 4 ways.

$\therefore$ Required number of ways $= 4 \times 4 \times 4 \times 4 \times 4 $
$ = 4^{5} = 1024$

13. How many words can be formed with the letters of the words COMMITTEE?

Solution:

The word COMMITTEE contains 9 letters, of which there are 2 ‘M’s, 2 ‘T’s, 2 ‘E’s.

$\therefore$ Required number of words $= \frac{\lfloor 9}{\lfloor 2 \times \lfloor 2 \times \lfloor 2} $
$ = \frac{362880}{2 \times 2 \times 2} $
$ = \frac{362880}{8} $
$ = 45360$

14. How many ways 10 hats of different colours can be put on the heads of 10 boys sitting in a row, so that the red and the black coloured hats can never be put on the heads of any two adjacent boys?

Solution:

Total number of ways 10 hats can be arranged in $ {^{10}P_{10}} = \lfloor 10$ ways.

Now, consider the red and black hats together as a single unit.

Then there are $8 + 1 = 9$ units to distribute among the 10 positions, which can be done in $\lfloor 9$ ways.

Again, the red and black hats can be arranged among themselves in $\lfloor 2$ ways.

So, total number of ways 10 hats can be arranged such that the red and black hats are always adjacent $= \lfloor 9 \times \lfloor 2$.

$\therefore$ Required number of ways $= \lfloor 10 – \lfloor 9 \times \lfloor 2 $
$ = 10 \times \lfloor 9 – 2 \times \lfloor 9 $
$ = (10 – 2)\lfloor 9 $
$ = 8 \times \lfloor 9 $

15. Find the value of n if:

(i) $2 \times {^{9}P_{n}} = {^{10}P_{n}}$

Solution:

$2 \times \frac{\lfloor 9}{\lfloor 9-n} = \frac{\lfloor 10}{\lfloor 10-n}$

$\Rightarrow \frac{2 \times \lfloor 9}{\lfloor 9-n} = \frac{10 \times \lfloor 9}{(10-n)\lfloor 9-n}$

$\Rightarrow 2 = \frac{10}{10-n}$

$\Rightarrow 2(10-n) = 10$

$\Rightarrow 20 – 2n = 10$

$\Rightarrow 2n = 10$

$\Rightarrow n = 5$

(ii) $^{n}P_{5} : {^{n}P_{3}} = 2 : 1$

Solution:

$\frac{^{n}P_{5}}{^{n}P_{3}} = \frac{2}{1}$

$\Rightarrow \frac{\frac{\lfloor n}{\lfloor n-5}}{\frac{\lfloor n}{\lfloor n-3}} = 2$

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

$\implies \frac{(n – 3)(n – 4)\lfloor{n – 5}}{\lfloor{n – 5}} = 2$

$\implies (n – 3)(n – 4) = 2$

$\implies n(n – 4) – 3(n – 4) = 2$

$\implies n^{2} – 4n – 3n + 12 – 2 = 0$

$\implies n^{2} – 7n + 10 = 0$

$\implies n^{2} – (5 + 2)n + 10 = 0$

$\implies n^{2} – 5n – 2n + 10 = 0$

$\implies n(n – 5) – 2(n – 5) = 0$

$\implies (n – 5)(n – 2) = 0$

Since $^{n}P_{5}$ is defined only for $n \ge 5$, $n = 2$ is not acceptable.

$\therefore$ Required value of $n = 5$

(iii) $3 \times {^{n}P_{4}} = 5 \times {^{n-1}P_{4}}, \quad n > 4$

Solution:

$3 \times {^{n}P_{4}} = 5 \times {^{n-1}P_{4}}, \quad n > 4$

$\Rightarrow 3 \times \frac{\lfloor n}{\lfloor n-4} = 5 \times \frac{\lfloor n-1}{\lfloor (n-1)-4}$

$\Rightarrow 3 \times \frac{\lfloor n}{\lfloor n-4} = 5 \times \frac{\lfloor n-1}{\lfloor n-1-4}$

$\Rightarrow 3 \times \frac{\lfloor n}{\lfloor n-4} = 5 \times \frac{\lfloor n-1}{\lfloor n-5}$

$\Rightarrow 3 \times \frac{n\lfloor n-1}{\lfloor n-4} = 5 \times \frac{\lfloor n-1}{\lfloor n-5}$

$\Rightarrow \frac{3n}{(n-4)\lfloor n-5} = \frac{5}{\lfloor n-5}$

$\Rightarrow \frac{3n}{n-4} = 5$

$\Rightarrow 3n = 5(n-4)$

$\Rightarrow 3n = 5n – 20$

$\Rightarrow 2n = 20$

$\Rightarrow n = 10$

(iv) $^{n}P_{5} = 42 \times {^{n}P_{3}}, \quad n > 4$

Solution:

$\implies \frac{\lfloor{n}}{\lfloor{n – 5}} = 42 \times \frac{\lfloor{n}}{\lfloor{n – 3}}$

$\implies \frac{\lfloor{n}}{\lfloor{n – 5}} = 42 \times \frac{\lfloor{n}}{(n – 3)(n – 4)\lfloor{n – 5}}$

$\implies 1 = \frac{42}{(n – 3)(n – 4)}$

$\implies (n – 3)(n – 4) = 42$

$\implies n(n – 4) – 3(n – 4) = 42$

$\implies n^{2} – 4n – 3n + 12 – 42 = 0$

$\implies n^{2} – 7n – 30 = 0$

$\implies n^{2} – (10 – 3)n – 30 = 0$

$\implies n^{2} – 10n + 3n – 30 = 0$

$\implies n(n – 10) + 3(n – 10) = 0$

$\implies (n – 10)(n + 3) = 0$

$\therefore$ Required value of $n = 10$