Question 7 - Jee advanced Math 2022 P1 Questions with Solutions

The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is _____.

Sol :

Method 1 (long)

Let a, b, c and d be the four digits of a number from left to right in the given interval [2022, 4482].

\(\underline{a} \; \underline{b} \; \underline{c} \; \underline{d}\)

For a four digit number in [2022, 4482], the first digit ‘a’ is always 2 or 3 or 4. All three of them are present in the given set of digits. So ‘a’ can be chosen in 3 ways. 

But the choice for the second digit ‘b’ will depend on the choice of ‘a’. For instance, if ‘a’ is 3, ‘b’ can be any of the 6 given numbers(6 ways), but if ‘a’ is 4, ‘b’ cannot be 6 and 7 as the numbers 46_ _  and 47_ _are not a part of the interval. So if ‘a’ is 4, possible values of b are 0, 2, 3, 4(4 ways). 

So to make this a little easier, we divide [2022, 4482] into union of three smaller intervals(for 3 different choices of ‘a’) as follows :

[2022, 2999] U [3000, 3999] U [4000, 4482]

In [2022, 2999], first digit ‘a’ can only be 2(1 way). Second digit ‘b’ can be any of the 6 given numbers(6 ways). But the choice of third digit ‘c’ will depend on the choice of ‘b’. If ‘b’ is 0, ‘c’ cannot be 0 as the numbers 200_ are not a part of the interval(5 ways). For any value of ‘b’ other than 0, ‘c’ can be any of the six given numbers(6 ways). 

So we further divide the interval as follows:

[2022, 2099] U [2100, 2999] U [3000, 3999] U [4000, 4482]

In [2022, 2099], ‘a’  and ‘b’ can be chosen in 1 way each. Third digit ‘c’ can be any of 2, 3, 4, 6, 7, and hence can be chosen in 5 ways. But the choice of fourth digit ‘d’ will depend on ‘c’. If ‘c’ is 2, ‘d’ cannot be 0, so there are only 5 choices for ‘d’ (5 ways). But for any other value of ‘c’, ‘d’ can be any of 6 given numbers(6 ways). 

Hence, the interval will be further divided into

[2022, 2029] U [2030, 2099] U [2100, 2999] U [3000, 3999] U [4000, 4482]

And now our task simplifies a little. In [2022, 2029], ‘a’, ‘b’ and ‘c’ can be chosen in 1 way each.  And the fourth digit ‘d’ can be chosen in 5 ways.

\(\implies\) all four digits can be chosen in \(= 1 \times 1 \times 1 \times 5\) ways
                                                                \(= 5\)

In [2030, 2099], ‘a’ can be chosen in 1 way, ‘b’ also in 1 way, ‘c’ in 4 ways and ‘d’ in 6 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 1 \times 4 \times 6\) ways
                                                                \(= 24\)

In [2100, 2999], ‘a’ can be chosen in 1 way, ‘b’ in 5 ways, ‘c’ in 6 ways and ‘d’ in 6 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 5 \times 6 \times 6\) ways
                                                                \(= 180\)

In [3000, 3999], ‘a’ can be chosen in 1 way, ‘b’ in 6 ways, ‘c’ in 6 ways and ‘d’ in 6 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 6 \times 6 \times 6\) ways
                                                                \(= 216\)

But [4000, 4482] will again cause slight issues. We will need to divide it into [4000, 4399] U [4400, 4479] U [4480, 4482].

In [4000, 4399], ‘a’ can be chosen in 1 way, ‘b’ in 3 ways, ‘c’ in 6 ways and ‘d’ in 6 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 3 \times 6 \times 6\) ways
                                                                \(= 108\)

In [4400, 4479], ‘a’ can be chosen in 1 way, ‘b’ also in 1 way, ‘c’ in 6 ways and ‘d’ in 6 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 1 \times 6 \times 6\) ways
                                                                \(= 36\)

In [4480, 4482], ‘a’ can be chosen in 1 way, ‘b’ also in 1 way, ‘c’ in 0 ways and ‘d’ in 2 ways. 

\(\implies\) all four digits can be chosen in \(= 1 \times 1 \times 0 \times 2\) ways
                                                                \(= 0\)

Therefore, in the entirety of the interval [2022. 4482],

all four digits can be chosen in \(5 + 24 + 180 + 216 + 108 + 36 + 0\) ways
                                                                \(= 569\)

Therefore, 569 total numbers belonging to [2022, 4482] can be formed from the given set of digits.
————————

Method 2 (short)

Let a, b, c and d be the four digits of a number from left to right in the given interval [2000, 4999].

\(\underline{a} \; \underline{b} \; \underline{c} \; \underline{d}\)

In the interval [2000, 4999], ‘a’ can be assigned a value of 2 or 3 or 4 from the six given digits, so ‘a’ can be chosen in 3 ways. Each of b, c and d can be assigned any of the six values(0, 2, 3, 4, 6, 7). This interval is cleverly chosen so that the choice for any digit do not depend on the choices of the other three digits. Meaning, regardless of the choices for ‘a’, ‘b’ and ‘c’, the fourth digit ‘d’ can be assigned any of the six given values and the resulting 4-digit number will be in this interval only. 

So ‘b’ can be chosen in 6 ways; ‘c’ in 6 ways; and ‘d’ in 6 ways. So, total numbers that can be formed in this new enhanced interval is \(3 \times 6 \times 6 \times 6 = 648\) ways.

But we want the numbers to belong to the interval [2022, 4482] instead. So from the above 648 total numbers formed, we will subtract number of four-digit numbers that are less than 2021, i.e., between 2000 and 2022. And also number of four-digit numbers that are greater than 4482, i.e., between 4483 and 4999. 

For the numbers between 2000 and 2021, first and second digits are always 2 and 0. So ‘a’ and ‘b’ can be chosen in 1 way each. The third digit ‘c’ can be either 0 or 2. If ‘c’ is 0, then ‘d’ can be any of the six given digits(6 ways). And if ‘c’ is 2, ‘d’ can only be 0(1 way). Therefore, 6 + 1 = 7 numbers can be formed between 2000 and 2022.

Similarly, for the numbers between 4483 and 4999, first digit is always 4. So ‘a’ can be chosen in 1 way. Second digit ‘b’ can be 4 or 6 or 7. If ‘b’ is 6, ‘c’ can be any of the 6 given numbers and ‘d’ can be any of the 6 given numbers(6 \(\times\) 6 = 36 ways). Similarly, if ‘b’ is 7, there are 36 ways of selecting both ‘c’ and ‘d’. However, if ‘b’ is 4, ‘c’ can be either 8 or 9. But neither of them are in the given set of numbers. So 0 ways to chose ‘c’ when ‘b’ is 4.
\(\implies 36 + 36 + 0 = 72\) numbers between 4483 and 4999.

\(\implies\) total 4-digit numbers that can be formed in [2022, 4482] from given digits = 648 - (72 + 7) = 569.


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