How many natural nos. not exceeding 4321 can be formed with digits : 4, 3, 2, and 1 if digits can repeat?

Question

D answer is ... "313"... bt i dunno how 2 go abt it.. help...related to permutations n combinations

hey... garypopkins rocks,... he's a genius!!!...look at d way he's explained it.. .better dan even my maths teacher i bet!!!
thanx dude... and yeah agree with whatever u said abt oders ....!!! lol

Answer 1

As you can see, the answers so far are useless.

Summer forgot that you can repeat digits.

Pragyp got the same answer but only God knows how. Pragyp thinks that a number with 4 in the hundreds' place is too big, such as 3,421. Go figure.

Let's first see how many numbers can be formed from the four digits, and then throw out the ones that are too big.

You didn't say they had to be four-digit numbers.

Let's start with the one-digit numbers. You can have four of them.

How about two-digit numbers? You can have 16 of them, with repeating.

Three-digit numbers. You can have 4 * 4 * 4, or 64, with repeating.

As for the four-digit numbers,
You can have in the thousands' place 1, 2, 3, or 4.
You can have in the hundreds' place 1, 2, 3, or 4.
etc.

The number of four-digit numbers from 1, 2, 3, and 4, with repeating, is

4 * 4 * 4 * 4 = 256

This is a total of 340. Some are too big and have to be thrown out.

All four-digit numbers that begin with 44xx have to go. There are 16 of those. All four-digit numbers that begin with 43xx, and have 3 or 4 in their tens' place, are too big and must go. There are 8 of those.

We are left with some numbers that begin with 432x. Three of those are too big.

There you have it:
340 gross - 16 - 8 - 3 = 313

Math_kp has some nerve saying my answer is not correct. Math_kp agrees with me about throwing out 16, 8, and 3 numbers, but Math_kp forgot about the one-, two-, and three-digit numbers, and gets the wrong answer, too small by 84. 84 is the total number of one-, two-, and three-digit numbers.

Answer 2

So far this question has not ben answered properly.

if the number is not > 4321 then we can count the total number of numbers and subtract the number > 4321

in any position there is 4 choices so total number of numbers = 4*4*4*4 = 256
now find the numbers that should be dropped

1) 1st 2 digit 4 3rd and 4th can be any thing 16 choices
2) 1st 4 2nd 3 third 3 or 4 4th any thing 8 choices
3) 1st 4 2nd 3 rd 2 4th 3 choices

number of number >4321 = 16+8+ 3 = 27

so number of numbers <= 4321 = 256-27 = 229

Answer 3

There are 24 different variations. I cannot rememer the mathemateical term, but if you multiply 4x3x2x1, then you will get the amount of variations. In math, you may see it as "4!" (four with the exclamation point on the end). You always do that when you try to figure out how many variations there are in a numbered situation as such. so if you have something with 12 digits and you want to figure out how many different numbers you can come up with, you myltiply 12x11x10x9x8x7x6x5x4x3x2x1 and you get the answer!

now, if you are trying to see how many you can get with NOT using all of the numbers, you would do the same thing..... just with three variations, two wariations, and one variation......

Hope this helps!

The answers (using all four numbers all of the time):
1234
1243
1342
1324
1423
1432
2134
2143
2314
1341
2431
2413
3124
3142
3241
3241
3412
3421
4321
4312
4213
4231
4123
4132

Answer 4

at thousands place you can have any of 1,2,3,4. hence you have 4 possibilities
at hundreds place you can have 1 or 2 or 3 but not 4 as it will exceed 4321.so you have 3 possibilities
at ten's place you can have 2 or 1. so you have 2 possibilities
at unit's place you can have only 1.so you have 2 possibility.
so total you have 4*3*2*1=24 possibilities NOT 313.
so you can have 24 numbers possible out of 1,2,3,4 which are not exceeding 4321.