Question

a
During the math class, a teacher clears the concept of permutation and combination to the 11th standard students. After the
class was over she asks the students some questions, one of the question was: how many numbers between 99 and 1000 (bothexcluding) can be formed such that
i. Every digit is either 3 or 7.
8 ways
d. 16 ways
b. 2 ways
c. 27 ways
ii.There is no restriction.
a.1000ways
b.900ways
c.800ways
d.700ways
iii. No digit is repeated.
a.684 ways
b. 600 ways
d. 729 ways
c. 648 ways
iv. The digit in hundred's place is 7.
a.70
b.80
c.90
d.100
At least one of the digits is 7.
a.252 ways
b. 525 ways
c. 200 ways
d. 500 ways
​

Answer 1

Answer:

a) 2ways

b) 900ways

c) 648ways

d) 100

e)100 ways

Answer 2

Answer:

Given

numbers between 99 and 1000 (both excluding)

So it means numbers from 100 to 999 ( three digit number)

1. every digit is either 3 or 7

So the number of ways to fill all the three places is

2x2x2= 8 ways

a) 8 ways

2. there is no restriction

to form a three digit number we can use any number in ones and tens place but for hundredth place we can use fro 1-9 but not 0

So number of ways is

9x10x10= 900ways

b) 900 ways

3. No digit is repeated

Means we can use single digit only once

So number of ways is

9x9x8= 648 ways

c) 648ways

4. The digit at hundred's place is 7.

Means the number starts with only 7

So number of ways is

1x10x10=100ways

d) 100 ways

5. At least one of the digits is 7.

at least one seven, first find the numbers that have exactly one 7, then the numbers that have two 7s, and then the number that has three 7s and then add the results.

To find the number of numbers, think of choosing a digit for each spot: _ _ _

For one seven, you can fix a 7 in a spot, say the first one, so the number looks like 7_ _ and note that for the other spots you can have any of the other 9 digits (0,1,2,3,4,5,6,8, or 9), so there are 9⋅9=81 such numbers.

For numbers of the forms _ 7 _ and _ _ 7 the count is different because the first digit cannot be 0, so there are 8⋅9=72 possibilities for each.

Thus, in total there are 72+72+81=225 three-digit positive integers with one seven as a digit.

Two sevens: For the form _77 there are 8 possibilities because the first spot cannot be 0, and for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are 8+9+9=26 three-digit positive integers with one seven as a digit.

Three sevens: There is only one, 777.

So in total there are 225+26+1=252 three-digit integers with a seven as a digit.

a) 252 ways