What's the perimeter using Pythagorean theorem
1 answer:
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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 = = <em>294 </em>
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<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 = = <em>180</em>
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<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 = = <em>75</em>
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<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 = = 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 = = 21
Total number of required ways = 147 + 21 = <em>168</em>
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<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 = = 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 = = 15
Total number of required ways = 90 + 15 = <em>105</em>