Permutation Combination

Question

How many natural number not exceeding 4321 can be formed with the digits 1, 2, 3, 4, if the digits can be repeated?

Comments

no. of natural numbers not exceeding 4321 :

if the no. is 1 digit

  no. of ways = 4

OR 

if the no. is 2 digit

no. of ways = 4 * 4 = 16  [ as the digits can be repeated]

OR 

if the no. is 3 digit

no. of ways = 4 * 4 * 4 = 64

OR 

if the no. is 4 digit

no. of ways = 229

So, total no. of ways will be = 4 + 16 + 64 + 229 = 313

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yes,

Another thing is missing in ans,

Have u noticed

that we have taken started with 4

started with 3

But where are the cases where starting with 1 or 2??

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@srestha i think in the answer @sukanya had taken the 1st case as if the numbers were starting with 1 or 2 or 3

Answer 1

1. Starting with 1/2/3
   3      *      4       *      4       *   4          = 192

-----          -----          -----       -------
1/2/3     1/2/3/4      1/2/3/4    1/2/3/4
OR
2. Starting with 4
  i) starting with 4 and 1/2
          1    *     2       *     4        *   4            = 32
     --------     ------        ------      -------
          4        1/2         1/2/3/4      1/2/3/4
  OR
  ii) starting with 43
      a) starting with 431
           1   *   1  *  1  *   4                 = 4
         ----     ---    ---    ------
          4       3      1     1/2/3/4
      OR
      b) starting with 432
           1  *   1  *  1   * 1           = 1
         ----    ----   ----   ----
           4      3      2     1
 ----------------------------------------------------------------            
  Total no. of ways = 192 + 32 + 4 + 1
                                = 229 ways