6 * (digits) and 3 ||| --> ******||| --> # of permutations of these symbols is \(\frac{9!}{6!3!}\).

Or: The total number of ways of dividing n identical items (6 *'s in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 6 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}\).

Hope it's clear.

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

(A) 31
(B) 51
(C) 56
(D) 62
(E) 93

there is a shortcut. For the problem, 4 digits are equally important in 0000-9999 set and it is impossible to build a number using only one digit (like 11111) So, answer has to be divisible by 4. Only 56 works.

Posted from GMAT ToolKit

There are numbers of the form dd|d|dd| = 8C3 = 56

IMO C

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

There are numbers of the form dd|d|dd| = 8C3 = 56

IMO C