Question

Read the case study given below and attempt any4 sub-parts:
During themathematicsclass, a teacher clears the concept of permutation and combination to the 11" standard students. After the class he asks the students some questions, one of the question was in how many_ways numbers between 99 and 1000 (both excluding) can be formed such that

1. every digit is either 3 or 7

(a) 8 ways
(b) 2 ways
(c) 27 ways
(d) 16 ways

2. there is no restriction

(a) 1000 ways
(b) 900 ways
(c) 800 ways
(d) 700 ways

3. No digit is repeated

(a) 684 ways
(b) 600 ways
(c) 648 ways
(d) 729 ways

4. The digit at hundred'splace is 7.

(a) 70 ways
(b) 80 ways
(c) 90 ways
(d) 100 ways

5. The digit at the unit's place is 0.

(a) 252 ways
(b) 100 ways
(c) 90 ways
(d) 80 ways​

Answer 1

Answer:

1-d

2-b

3-a

4-c

5-d

ok. please follow

Answer 2

Given,

The numbers between 99 and 1000 are given.

To find,

We have to find the number of ways

(1)every digit is either 3 or 7

(2) there is no restriction

(3) No digit is repeated

(4) The digit at hundreds of places is 7.

(5)The digit at the unit's place is 0.

Solution,

The required answers are 1. (a) 8 ways,  2. (b) 900 ways, 3. (c) 648 ways, 4. (d) 100 ways, 5. (c) 90 ways.

Numbers between 99 and 1000 both exclude means from 100 to 999 i.e. 3-digit numbers.

(1) Fix the three positions with 3 or 7, then the number of ways that every digit is either 3 or 7 is 2*2*2 = 8

⇒ (a) 8 ways

(2) Now,  there is no restriction that means we can use any number to form 3-digit numbers, we cannot put 0 at hundredth place otherwise the number will become 2-digit, so the number of ways of forming 3-digit numbers is

              = 9*10*10

              = 900

⇒ (b) 900 ways

(3) Now, no digit is repeated, meaning we can use a single digit only one time, so the number of ways is

           = 9*9*8

           = 648 ways

⇒ (c) 648 ways

(4) Now, the digit at hundreds of places is 7, fix 7 at the hundredth place, number of ways are

          = 1*10*10

         = 100 ways

⇒(d) 100 ways

(5) Now, the digit at the unit's place is 0, fixing 0 at one place,

number of ways = 9*10*1

                           = 90 ways

⇒(c) 90 ways

Hence, the required answers are 1. (a) 8 ways,  2.(b) 900 ways, 3. (c) 648 ways, 4. (d) 100 ways, 5. (c) 90 ways.