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The number of $4$ digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set $\{1, 2, 3\}$ is ________.
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3 votes
we have 3 digits 1,2,3 and 4  places to form a number with repetitions of course.

No of r combinations from n distinct objects with repetitions  = $\binom{n+r-1}{r}$
 

 

Here r=4, n=3 and every combination will give only one number because we have the restriction that numbers are in non decreasing order.

Thus the answer is  $\binom{4+3-1}{4}$

= 15
2 votes
2 votes

This can be visualized as Building a tree in such a way that child is either equal or greater than the parent.

Find the number of leafs in such a tree of 4 levels.

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1 votes

One way of solving

S = {3,2,1}

all allowed ordered pairs

  1. (1,1)
  2. (1,2)
  3. (1,3)
  4. (2,2)
  5. (2,3)
  6. (3,3)

we can append all 6 pairs after (1,1) = 6 (1111,1112,1113,1122,1123,1133)
we can append all pairs after (1,2) with (1,2) except 3rd = 3
we can append all pairs after (1,3) with (1,2) except 3rd = 2
we can append 4th 5th and 6th pairs with (2,2) = 3
we can append only one pair 6th with (2,3) = 1
we can append (3,3) with itself  = 1

6+3+2+3+1+1 = 15

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0 votes
Just try......
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4-digit no. are possible.
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