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# how many natural numbers not exceeding 4321can be formed with the digits 1,2,3,4 if the digits can repeat?

sanatan sharma
26 Points
9 years ago

sarina k
18 Points
9 years ago

Anyway, there are 4 one digit numbers (obvious)
16 two digit numbers (there are 4 options for each digit, and two digit numbers are automatically <= 4321)
64 three digit numbers (again, 4 options for each digit)

Now for the four digit numbers:
There are 64 numbers where the first digit is a 3 (4 options for each of the remaining digits, and since the first digit is a 3, the number is < 4321)

Similarly there are 64 numbers where the first digit is a 2, and another 64 where the first digit is a 1.
This is 192 four digit numbers so long.

Now lets look at what happens if the first digit is a 4.
There are 16 numbers where the second digit is a 2 (the last two digits can be anything in this case) and another 16 where the second digit is a 1.
If the second digit is a 3:
...If the third digit is a 1, there are 4 numbers, because there are no restrictions on the last digit.
...If the third digit is a 2, there is only one possible number, n.l. 4321.

So, there are 16 numbers if the second digit is a 2, 16 numbers if the second digit is a 1, and 4 + 1 = 5 numbers if the second digit is a 3, giving
16 + 16 + 5 = 37 four digit numbers where the first digit is a 4.

In total, we have
4 one digit numbers
16 two digit numbers
64 three digit numbers
192 four digit numbers where the first digit is not a 4
37 four digit numbers where the first digit is a 4

This is 313 numbers in total.

KAMAL RAWAT
39 Points
9 years ago

answer is 4 raise to power 4.

Priyanka
11 Points
3 years ago
No. has to be less than 4321
so, consider no. s less than 4000

1st place:3 ways
2nd place:4 ways
3rd palce:4 ways
4th place:4 ways
therefore, 3*4*4*4=192

now consider no.s starting with 4
2nd place can be filled in 2 ways (1&2) as no. has to be less than 4321
1st place:1 way
2nd place:2 ways
3rd palce:4 ways
4th place:4 ways
therefore,1*2*4*4=32

now consider no.s starting with 4 and followed by 3
case 1: 432_ → 1 way
case 2:431_ → 4 ways
therfore , 1+4= 5 ways

So , total no. of numbers formed= 192+32+5=229
Anas
20 Points
3 years ago
Anweres is When first two digits are fixed that is 4,4 Now when two are fixed from satrting that is 4,3=16+8+3=27Total ways =340SoNo.`s less than 4321=340-27=319
Anas
20 Points
3 years ago
Total no. Of ways = 4*4*4*4+4*4*4+4*4+4=340No.`sgreater than 4321Fixing first two no.`s i.e.,4,4,_,_No. Of ways ,n1=4*4=16Fixing first two no`s i.e.,4,3,_,_No.of ways,n2 =2*4=8Fixing first three no.`s i.e.,4,3,2,_No.of ways,n3=3Now No.`ss less than 4321are=T-(n1+n2+n3)=340-27=313
Rishi Sharma
one year ago
Dear Student,

there are 4 one digit numbers (obvious)
16 two digit numbers (there are 4 options for each digit, and two digit numbers are automatically <= 4321)
64 three digit numbers (again, 4 options for each digit)
Now for the four digit numbers:
consider no. s less than 4000
1stplace:3 ways
2ndplace:4 ways
3rdpalce:4 ways
4thplace:4 ways
therefore, 3*4*4*4=192
now consider no.s starting with 4
2ndplace can be filled in 2 ways (1&2) as no. has to be less than 4321
1stplace:1 way
2nd place:2 ways
3rdpalce:4 ways
4thplace:4 ways
therefore,1*2*4*4=32
now consider no.s starting with 4 and followed by 3
case 1: 432_ → 1 way
case 2:431_ → 4 ways
therfore , 1+4= 5 ways

So , total no. of 4 digit numbers formed = 192+32+5=229
hence total number = 4+16+64+229 = 313

Thanks and Regards