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MATHS 5 Marks Questions

Permutation and Combinations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 4 questions.

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Question 15 Marks
Match each item given under the column C1 to its correct answer given under the column C2.
Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
C1
C2
(a)
Boys and girls alternate.
(i)
5! × 6!
(b)
No two girls sit together.
(ii)
10! – 5! 6!
(c)
All the girls sit together.
(iii)
(5!)2 + (5!)2
(d)
All the girls are never together.
(iv)
2! 5! 5!
Answer
 C1
C2
(a)
 Boys and girls alternate.
(iii)
 (5!)2 + (5!)2
(b)
 No two girls sit together.
(i)
 5! × 6!
(c)
 All the girls sit together.
(iv)
 2! 5! 5!
(d) All the girls are never together. (ii) 10! – 5! 6!
Explanation:
Total number of arrangment when boys and girls alternate: = (5!)2 + (5!)2
  1. No two girls sit together: = 5! 6!
  2. All the girls sit never toghether = 2! 5! 5!
  3. All the girls sit never together = 10! - 5! 6!
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Question 25 Marks
Match each item given under the column C1 to its correct answer given under the column C2.
Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find
C1
C2
(a)
how many numbers are formed.
(i)
840
(b)
how many numbers are exactly divisible by 2.
(ii)
200
(c)
how many numbers are exactly divisible by 25.
(iii)
360
(d)
how many of these are exactly divisble by 4.
(iv)
40
Answer
C1
C2
(a)
how many numbers are formed.
(i)
840
(b)
how many numbers are exactly divisible by 2.
(iii)
360
(c)
how many numbers are exactly divisible by 25.
(iv)
40
(d)
how many of these are exactly divisble by 4.
(ii)
200
Explanation:
  1. Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7 $=\ ^7\text{P}_4=\frac{7!}{(7-4)!}=\frac{7\times6\times5\times4\times3!}{3!}=840$
  2. When anumber is divisible by $2=4\times5\times6\times3=360$
  3. Total number which are divisible by 25 = 40
  4. Total number which are divisible by 4 (last two digits is divisible by 4) = 200
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Question 35 Marks
Match each item given under the column C1 to its correct answer given under the column C2.
There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturer is to be formed. Find:
C1
C2
(a)
In how many ways committee can be formed.
(i)
10C2 × 19C3
(b)
In how many ways a particular professor is included.
(ii)
10C2 × 19C2
(c)
In how many ways a particular lecturer is included.
(iii)
9C1 × 20C3
(d)
In how many ways a particular lecturer is excluded.
(iv)
10C2 × 20C3
Answer
C1
C2
(a)
 In how many ways committee can be formed.
(iv)
 10C2 × 20C3
(b)
 In how many ways a particular professor is included.
(iii)
 9C1 × 20C3
(c)
 In how many ways a particular lecturer is included.
(ii)
 10C2 × 19C2
(d) In how many ways a particular lecturer is excluded. (i) 10C2 × 19C3
Explanation:
  1. We have to select 2 professor out of 10 and 3 lecturers out of 20 $\therefore$ Number of ways of selection = 10C2 × 20C3
  2. When a paeticular professor is included the number of ways = 10 - 1C1 × 20C3 = 9C1 × 20C3
  3. When a particular lecturer is included number of ways = 10C2 × 19C2
  4. Whan a particular lecturer is excluded, then number of ways = 10C2 × 19C3
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Question 45 Marks
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Answer
We have to form 4-digit numbers which are greater than 6000 and less than 7000.
We know that a number is divisible by 5, if at the unit place of the number there is 0 or 5.
So, unit digit can be filled in 2 ways.
The thousandth place can be filled by ‘6’ only.
The hundredth place and tenth place can be filled together in 8 × 7 = 56 ways. So, total number of ways = 56 × 2 = 112
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