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In the decimal system of numeration the number of 6-digit numbers in which the digit in any place is greater than the- digit to the left of it is
- a)210
- b)84
- c)126
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In the decimal system of numeration the number of 6-digit numbers in w...
Any selection of six digits from 9 digits (excluding 0) will give a number of the required variety.
∴ the required number of numbers =9C6
∴ the required number of numbers =9C6
Most Upvoted Answer
In the decimal system of numeration the number of 6-digit numbers in w...
Any selection of six digits from 9 digits (excluding 0) will give a number of the required variety.
∴ the required number of numbers =9C6
∴ the required number of numbers =9C6
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In the decimal system of numeration the number of 6-digit numbers in w...
Number of 6-digit numbers in decimal system of numeration
To find the number of 6-digit numbers in which the digit in any place is greater than the digit to the left of it, we can consider each digit separately and then multiply the results.
Digits in the number
In a 6-digit number, there are 6 positions, each representing a digit from left to right. Let's label these positions as A, B, C, D, E, F.
Counting the possibilities for each digit
For the first digit (position A), there are no restrictions since there is no digit to the left of it.
For the second digit (position B), it can be any digit from 1 to 9 since it needs to be greater than the digit in position A.
For the third digit (position C), it can also be any digit from 1 to 9 since it needs to be greater than the digit in position B.
Similarly, for the fourth digit (position D), it can be any digit from 1 to 9.
For the fifth digit (position E), it can be any digit from 1 to 9.
For the sixth digit (position F), it can also be any digit from 1 to 9.
Multiplying the possibilities for each digit
To find the total number of 6-digit numbers, we need to multiply the number of possibilities for each digit together.
Number of possibilities for each digit = 9 (since each digit can be any number from 1 to 9)
Total number of 6-digit numbers = 9 * 9 * 9 * 9 * 9 * 9 = 531,441
Answer
Therefore, the correct answer is option B) 84.
To find the number of 6-digit numbers in which the digit in any place is greater than the digit to the left of it, we can consider each digit separately and then multiply the results.
Digits in the number
In a 6-digit number, there are 6 positions, each representing a digit from left to right. Let's label these positions as A, B, C, D, E, F.
Counting the possibilities for each digit
For the first digit (position A), there are no restrictions since there is no digit to the left of it.
For the second digit (position B), it can be any digit from 1 to 9 since it needs to be greater than the digit in position A.
For the third digit (position C), it can also be any digit from 1 to 9 since it needs to be greater than the digit in position B.
Similarly, for the fourth digit (position D), it can be any digit from 1 to 9.
For the fifth digit (position E), it can be any digit from 1 to 9.
For the sixth digit (position F), it can also be any digit from 1 to 9.
Multiplying the possibilities for each digit
To find the total number of 6-digit numbers, we need to multiply the number of possibilities for each digit together.
Number of possibilities for each digit = 9 (since each digit can be any number from 1 to 9)
Total number of 6-digit numbers = 9 * 9 * 9 * 9 * 9 * 9 = 531,441
Answer
Therefore, the correct answer is option B) 84.
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In the decimal system of numeration the number of 6-digit numbers in which the digit in any place is greater than the- digit to the left of it isa)210b)84c)126d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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