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belka [17]
3 years ago
12

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers

can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?6x6x5=180 c) How many odd numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?3x5x5=75 d) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?3x7x7=147/1x3x7=21/147+27=168 e) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?1x3x5=15/3x6x5=90/90+15=105
Mathematics
1 answer:
love history [14]3 years ago
7 0

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = 7 \times 7 \times 6 = <em>294 </em>

<em></em>

<em>Part B:</em>

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 6 \times 6 \times 5 = <em>180</em>

<em></em>

<em>Part C)</em>

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 3 \times 5 \times 5 = <em>75</em>

<em></em>

<em>Part d)</em>

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = 3 \times 7 \times 7 = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = 1 \times 3 \times 7 = 21

Total number of required ways = 147 + 21 = <em>168</em>

<em></em>

<em>Part e)</em>

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = 3 \times 6 \times 5 = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = 1 \times 3 \times 5 = 15

Total number of required ways = 90 + 15 = <em>105</em>

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