
The number of four digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2,3,4,5 (repetitions of digits are allowed) is:
A) 288
B) 306
C ) 360
D ) 310
Answer
573.6k+ views
Hint: Consider the different cases by putting the digits at different places. This will help you to get the correct answer.
Complete step by step answer:
We can see that we need to make a 4 digit number from 6 digits. Let us consider the cases.
Case- I:
If we selected the first digit as 5 and second, third and four digits can be any of the six digits.
\[\therefore \] Number of ways of selecting second, third and fourth digit = \[6\times 6\times 6=216\]
Case-II:
If we select the first digit as 4 then the second digit can take valves either 4 or 5.
\[\therefore \]Number of ways of selecting second digit =2
Again we can take any of the six digits for third and fourth digit.
\[\therefore \]Number of ways of selecting third and fourth digit = 6 ways each
Thus number of ways = \[2\times 6\times 6=72\]
Case-III:
Now, if we select first digit as 4, second digit as 3 and third digit - can only take the value 3,4,5 then number of ways =3
Fourth digit can take any value from 6 digits
\[\therefore \] Number of ways = 6
Thus number of ways = \[3\times 6=18\]
Case IV:
If we select first digit as 4, second digit as 3, third digit as 2, number of ways by which fourth digit can take the valves (2, 3, 4, 5) = 4
\[\therefore \] Number of ways = 4
Thus number of digit greater than 4321 = 216+72+18+4 = 310
So, the correct answer is “Option D”.
Note: The student must consider the cases in the correct way. Do not take the first digit as 0.
And the alternate short solution for this problem is:
The number of four-digit numbers Starting with 5 is equal to ${{6}^{3}}=216$.
Starting with 44 and 55 is equal to 36 × 2=72
Starting with 433, 434 and 435 is equal to 6 × 3=18
Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
216+72+18+4=310
Complete step by step answer:
We can see that we need to make a 4 digit number from 6 digits. Let us consider the cases.
Case- I:
If we selected the first digit as 5 and second, third and four digits can be any of the six digits.
\[\therefore \] Number of ways of selecting second, third and fourth digit = \[6\times 6\times 6=216\]
Case-II:
If we select the first digit as 4 then the second digit can take valves either 4 or 5.
\[\therefore \]Number of ways of selecting second digit =2
Again we can take any of the six digits for third and fourth digit.
\[\therefore \]Number of ways of selecting third and fourth digit = 6 ways each
Thus number of ways = \[2\times 6\times 6=72\]
Case-III:
Now, if we select first digit as 4, second digit as 3 and third digit - can only take the value 3,4,5 then number of ways =3
Fourth digit can take any value from 6 digits
\[\therefore \] Number of ways = 6
Thus number of ways = \[3\times 6=18\]
Case IV:
If we select first digit as 4, second digit as 3, third digit as 2, number of ways by which fourth digit can take the valves (2, 3, 4, 5) = 4
\[\therefore \] Number of ways = 4
Thus number of digit greater than 4321 = 216+72+18+4 = 310
So, the correct answer is “Option D”.
Note: The student must consider the cases in the correct way. Do not take the first digit as 0.
And the alternate short solution for this problem is:
The number of four-digit numbers Starting with 5 is equal to ${{6}^{3}}=216$.
Starting with 44 and 55 is equal to 36 × 2=72
Starting with 433, 434 and 435 is equal to 6 × 3=18
Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
216+72+18+4=310
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