How many natural numbers not exceeding 5432 can be formed with the digits 2, 3,4,5

There are actually $27$ numbers larger than $4321$. They are:

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How many numbers not exceeding 4321 can be formed using digits 1 2 3 4 if repetition of digits is allowed?
How many natural numbers can be formed using the digits 1 2 3 and 4?
How many natural numbers not more than 4300 can be formed with the digits?

Numbers of the form $44\square\square$: There are $1 \cdot 1 \cdot 4 \cdot 4 = 16$ such numbers.

Numbers of the form $433\square$ or $434\square$: There are $1 \cdot 1 \cdot 2 \cdot 4 = 8$ such numbers.

Numbers of the form $432\square$: There are $3$ such numbers, namely $4322, 4323, 4324$.

Hence, you should have $4^4 - (16 + 8 + 3) = 256 - 27 = 229$ in your first method.

In your second method, your forgot to count numbers in which two digits each appear twice, which is why your four cases do not add up to $4^4 = 256$. There are $\binom{4}{2}$ ways to select the two digits which each appear twice and $\binom{4}{2}$ ways to select two positions for the smaller of those digits. Hence, there are $$\binom{4}{2}\binom{4}{2} = 36$$ numbers which each appear twice.

Thus, you should have a total of $24 + 144 + 36 + 48 + 4 = 256$ numbers, of which $27$ are too large, again giving $256 - 27 = 229$ admissible numbers.

A direct count: We wish to count four-digit numbers formed using the digits $1, 2, 3, 4$ which are at most $4321$ when repetition of digits is permitted.

Numbers of the form $1\square\square\square$, $2\square\square\square$, or $3\square\square\square$: $3 \cdot 4 \cdot 4 \cdot 4 = 192$

Numbers of the form $41\square\square$ or $42\square\square$: $1 \cdot 2 \cdot 4 \cdot 4 = 32$

Numbers of the form $431\square$: $4$

The number $4321$: $1$

Hence, the numbers of admissible numbers is $192 + 32 + 4 + 1 = 229$.

(d) 313

As there are 4 digits 1, 2, 3, 4 and repetition of digits is allowed.

Total number of 1-digit numbers = 4

Total number of 2-digit numbers = 4 × 4 = 16

Total number of 3-digit numbers = 4 × 4 × 4 = 64

Number of 4-digit numbers beginning with 1 = 4 × 4 × 4 = 64

(∵ The first place is occupied by 1)

Number of 4-digit numbers beginning with 2 = 4 × 4 × 4 = 64

Number of 4-digit numbers beginning with 3 = 4 × 4 × 4 = 64

Number of 4-digit numbers beginning with 41 = 4 × 4 = 16

Number of 4-digit numbers beginning with 42 = 4 × 4 = 16

Number of 4-digit numbers beginning with 431 = 4

Number of 4-digit numbers beginning with 432 = 1 (4321 only)

∴ Total number of 4-digit numbers = 64 + 64 + 64 + 16 + 16 + 4 + 1 = 229

∴ Total number of natural numbers not exceeding 4321 = 4 + 16 + 64 + 229 = 313.

Case I: Four-digit number
Total number of ways in which the 4 digit number can be formed =`4xx4xx4xx4=256`

Now, the number of ways in which the 4-digit numbers greater than 4321 can be formed is as follows:
Suppose, the thousand's digit is 4 and hundred's digit is either 3 or 4.
∴ Number of ways =`2xx4xx4=32`

But 4311, 4312, 4313, 4314, 4321 (i.e. 5 numbers) are less than or equal to 4321.
∴ Remaining number of ways =`256-(32-5)=229`

Case II: Three-digit number
The hundred's digit can be filled in 4 ways.
Similarly, the ten's digit and the unit's digit can also be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of three-digit number =`4xx4xx4=64`

Case III: Two-digit number
The ten's digit and the unit's digit can be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of two digit numbers `4xx4=16`

Case IV: One-digit number
Single digit number can only be four.
∴ Required numbers = 229 + 64 + 16 +4 = 313

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How many four digit natural nu...

Updated On: 27-06-2022

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Text Solution

Solution : Total possibleways`=3*4^3+1+4+2*4^2`<br> `=192+5+32`<br> `=229`.

Answer

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