SEBA Class 10 Advanced Maths Chapter 6.1 Solutions: 

Here, you will get reliable SEBA Class 10 Advanced Maths Chapter 6.1 Solutions for the upcoming HSLC exam. In this article, we solved all the textual exercises of Chapter 6.1 Permutation and Combination from the SEBA Class 10 New Advanced Mathematics book. We provide all solutions of Chapter 6.1 in a very simple way.

1. Find the value of :

(i) $\lfloor 5$

$\lfloor 5 $
$ = 5 \times 4 \times 3 \times 2 \times 1 $
$ = 120$

(ii) $\lfloor 4 + \lfloor 5$

$\lfloor 4 + \lfloor 5 $
$ = \lfloor 4 + 5 \times \lfloor 4 $
$= (1 + 5)\lfloor 4 $
$= 6 \times 24 = 144$

(iii) $\frac{\lfloor 9}{\lfloor 7}$

$\frac{\lfloor 9}{\lfloor 7} $
$ = \frac{9 \times 8 \times \lfloor 7}{\lfloor 7} $
$= 9 \times 8 = 72$

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SEBA Class 10 Advanced Maths Chapter 4.1 Solutions

(iv) $\lfloor 6 – \lfloor 3$

$\lfloor 6 – \lfloor 3 $
$ = 6 \times 5 \times 4 \times \lfloor 3 – \lfloor 3 $
$= (120 – 1)\lfloor 3 $
$= 119 \times 6 = 714$

(v) $\frac{\lfloor 12}{\lfloor 2 \cdot \lfloor 10}$

$\frac{\lfloor 12}{\lfloor 2 \cdot \lfloor 10} $
$ = \frac{12 \times 11 \times \lfloor 10}{(2 \times 1) \times \lfloor 10} $
$= \frac{132}{2} = 66$

(vi) $\lfloor 3 \cdot \lfloor 7$

$\lfloor 3 \cdot \lfloor 7 $
$ = (3 \times 2 \times 1) \times (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) $
$= 6 \times 5040 $
$= 30240$

2. Show that, $\lfloor 2n = 2^n \cdot \lfloor n \{1 \cdot 3 \cdot 5 \cdots (2n-1)\}$

Proof:

L.H.S. $= \lfloor 2n$

$= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (2n-1) \cdot 2n$

$= \{1 \cdot 3 \cdot 5 \cdots (2n-1)\} \cdot \{2 \cdot 4 \cdot 6 \cdots 2n\}$

$= \{1 \cdot 3 \cdot 5 \cdots (2n-1)\} \cdot \{(2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdots (2 \cdot n)\}$

$= \{1 \cdot 3 \cdot 5 \cdots (2n-1)\} \cdot 2^n \cdot (1 \cdot 2 \cdot 3 \cdots n)$

$= 2^n \cdot \lfloor n \{1 \cdot 3 \cdot 5 \cdots (2n-1)\}$

$= \text{R.H.S.}$

$\therefore \lfloor 2n = 2^n \cdot \lfloor n \{1 \cdot 3 \cdot 5 \cdots (2n-1)\}$

Proved.

3. How many words (the words may or may not have meaning) can be formed with the letters of the word ENGLISH taking three at a time?

Solution:

The word ENGLISH contains $7$ distinct letters.

We need to form 3-letter words, which means filling up 3 vacant places (1st, 2nd, 3rd).

• The 1st place can be filled by any one of the $7$ letters in $7$ ways.

• The 2nd place can be filled from the remaining $6$ letters in $6$ ways.

• The 3rd place can be filled from the remaining $5$ letters in $5$ ways.

Hence, according to the fundamental principle of counting, the total number of words that can be formed is:

$= 7 \times 6 \times 5 = 210$

4. If repetition is allowed, how many even numbers of two digits can be formed with the digits 1, 2, 3, 4, 5?

Solution:

For the number to be even, the unit place must be filled with an even digit ($2$ or $4$).

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

• Since repetition is allowed, the tens place can be filled by any of the given $5$ digits in $5$ ways.

Hence, according to the fundamental principle of counting, the required number of two-digit even numbers is:

$= 5 \times 2 = 10$

5. Without repetition, how many numbers of 5 digits can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9?

Solution:

The given digits are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (total $10$ digits).

We have to form 5-digit numbers using 5 places: Ten-Thousands (1st), Thousands (2nd), Hundreds (3rd), Tens (4th), and Units (5th).

• The first place (ten-thousands place) cannot be filled with $0$, so it can be filled by any of the remaining $9$ digits ($1$ to $9$) in $9$ ways.

• Since repetition is not allowed and $0$ can now be used, the second place can be filled by any of the remaining $9$ digits in $9$ ways.

• The third place can be filled by any of the remaining $8$ digits in $8$ ways.

• The fourth place can be filled in $7$ ways.

• The fifth place can be filled in $6$ ways.

Hence, the total number of 5 digits can be formed with the given digits is :

$= 9 \times 9 \times 8 \times 7 \times 6 = 27216$

6. Without repetition how many 4 digit numbers can be formed with the digits 1, 3, 5, 7, 9?

Solution:

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

We have to fill 4 positions: Thousands (1st), Hundreds (2nd), Tens (3rd), and Units (4th).

• The 1st place can be filled by any of the $5$ digits in $5$ ways.

• The 2nd place can be filled by any of the remaining $4$ digits in $4$ ways.

• The 3rd place can be filled by any of the remaining $3$ digits in $3$ ways.

• The 4th place can be filled by any of the remaining $2$ digits in $2$ ways.

Hence, the total number of 4-digit numbers that can be formed with the given digits is:

$= 5 \times 4 \times 3 \times 2 = 120$

7. How many ways the letters of the word NUMBER can be arranged?

(i) How many of these will start with M?

(ii) How many of these do not end with B?

Solution:

The word NUMBER has $6$ distinct letters: N, U, M, B, E, R.

Total arrangements:

According to the fundamental principle of counting, the 6 letters can fill 6 positions in:

$= 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \lfloor 6 = 720\text{ ways}$

(i) Arrangements starting with M:

If M is fixed at the 1st position ( So it can be done in $1$ way).

The remaining 5 positions can be filled by the remaining 5 letters in:

$= 5 \times 4 \times 3 \times 2 \times 1 = \lfloor 5 = 120\text{ ways}$

$\therefore$ The number of arrangements starting with M $= 120$

(ii) Arrangements that do not end with B:

• If B is fixed at the last position. So the 6th place can be filled in $1$ way, and the remaining 5 places can be filled by the other 5 letters in:

  $= \lfloor 5 = 120\text{ ways}$

$\therefore$ Number of arrangements that do not end with B :

$= 720 – 120 = 600$