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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?

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

Grade:12

7 Answers

sanatan sharma
26 Points
9 years ago

the answer is 44 .

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
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

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

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