it is a finite set, repetition is not possible

1 vote

Number of 5 digit number having there digits in non decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3, 4, 5, 6, 7, 8, 9} ?

1 vote

5 digit no. with non decreasing order is same as selecting five digit and after selection we can arrange that 5 digit only in one way.

- if repetition is not allowed we can select in 9
**c**5. - and if repetition is allowed then we can select in 13
**c**5.

0

If repetition is not allowed we can select in 9C5 ways that is perfect when we given that digits are in strictly in increasing order.

But for repetition is allowed 13c5 how ???

But for repetition is allowed 13c5 how ???

0

if repetition is allowed then it is same as ** x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 5**

where 0 <=xi <=5 **. **It is the case of **5** indistinguishable objects into ** 9 **distinguishable boxes. $\binom{n+r-1}{r} = \binom{9+5-1}{5}$

0

why it is not the case where we have 5 boxes and 9 objects.

since in the question also we have 5 places to fill with these 9 objects so the ways can be calculated as

(9+5-1)C(5-1) = (9+5-1)C(9)

as according to combinatorics (boxes + object - 1) C (boxes -1) = (boxes + object -1) C (objects)

Correct me if I am wrong??

since in the question also we have 5 places to fill with these 9 objects so the ways can be calculated as

(9+5-1)C(5-1) = (9+5-1)C(9)

as according to combinatorics (boxes + object - 1) C (boxes -1) = (boxes + object -1) C (objects)

Correct me if I am wrong??

1

here we have x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 5 . for all xi ( box) we can leave the box vaccant ( xi = 0) and we can also take multiple value of every xi ( xi=1,2,3...). But the condition is total no of digit is 5.

But in your case x1+x2+x3+x4+x5=9 does not make sense. you take places as xi which can not be vaccant and also we can not take multiple value for xi (box) .

But in your case x1+x2+x3+x4+x5=9 does not make sense. you take places as xi which can not be vaccant and also we can not take multiple value for xi (box) .

0

ok, I got it since in digit places we can place only one digit at a time.

but in box and ball example we can put multiple balls at a time in a single box that's why it is calling as repetition and hence the number of boxes we have is 5 and objects as 9.

but in our example, we have 9 boxes(digits from the set) and 5 balls(places in five-digit number). Through this, we can take the same digit for multiple places. ryt???

but in box and ball example we can put multiple balls at a time in a single box that's why it is calling as repetition and hence the number of boxes we have is 5 and objects as 9.

but in our example, we have 9 boxes(digits from the set) and 5 balls(places in five-digit number). Through this, we can take the same digit for multiple places. ryt???