Out of the number of four different digits formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, how many of them are greater than 4000 (Repetition is not allowed)?
(a). 2010
(b). \[{}^9{P_4}\]
(c). 2016
(d). None of these
Answer
655.5k+ views
Hint: For the four-digit number to be greater than 4000, the first digit from the left should be equal to or greater than 4 and the remaining 3 digits should be chosen from the remaining digits.
Complete step-by-step answer:
The permutation is defined as each of several possible ways in which a set or number of things can be ordered or arranged.
The number of ways of arranging a thing is equal to the product of the number of them available to be arranged in a particular place.
We are given the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9. We need to form four different digits and find the number of them greater than 4000.
In the first digit from the left, the digits 4, 5, 6, 7, 8, 9 can be chosen making them a total of 6 possibilities.
In the second digit to the left, one number is already chosen, so we have 8 left.
In the third digit to the left, two numbers are already chosen, so we have 7 left.
In the fourth digit to the left, three numbers are already chosen and hence, we have 6 left.
Hence, the total number of ways is the product of the possibilities.
Total ways = \[6 \times 8 \times 7 \times 6\]
Total ways = 2016
Hence, the correct answer is option (c).
Note: You can also find the total number of four different digits greater than 4000 by selection of 4 digits from given numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and then subtract the number of four-digits less than 4000 which will give numbers greater than 4000 ((Repetition is not allowed).
Complete step-by-step answer:
The permutation is defined as each of several possible ways in which a set or number of things can be ordered or arranged.
The number of ways of arranging a thing is equal to the product of the number of them available to be arranged in a particular place.
We are given the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9. We need to form four different digits and find the number of them greater than 4000.
In the first digit from the left, the digits 4, 5, 6, 7, 8, 9 can be chosen making them a total of 6 possibilities.
In the second digit to the left, one number is already chosen, so we have 8 left.
In the third digit to the left, two numbers are already chosen, so we have 7 left.
In the fourth digit to the left, three numbers are already chosen and hence, we have 6 left.
Hence, the total number of ways is the product of the possibilities.
Total ways = \[6 \times 8 \times 7 \times 6\]
Total ways = 2016
Hence, the correct answer is option (c).
Note: You can also find the total number of four different digits greater than 4000 by selection of 4 digits from given numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and then subtract the number of four-digits less than 4000 which will give numbers greater than 4000 ((Repetition is not allowed).
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