parveen110 wrote:The total number of natural numbers of six digits that can be formed with digits 1,2,3,4, if all the digits are to appear in the same number at least once, is:
a. 740
b. 1830
c. 1560
d. none of these

Number of ways to arrange n distinct elements = n!.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Thus:
Number of ways to arrange AAABB = 5!/(3!2!).
We divide by 3! to account of the 3 identical A's and by 2! to account for the 2 identical B's.

Option A: 3 digits the same, the other 3 different
Case 1:
Let's say the 6-digit integer is to be composed of 1,1,1,2,3,4.
Number of ways to arrange the 6 digits = 6!/3! = 120.
Other cases:
The digit in red could be 1, 2, 3, or 4, for a total of 4 options.
Thus, we multiply by 4:
120*4 = 480.

Option B: 2 digits the same, another 2 digits the same, the other 2 different
Case 1:
Let's say the 6-digit integer is to be composed of 1,1,2,2,3,4.
Number of ways to arrange the 6 digits = 6!/(2!2!) = 180.
Other cases:
Case 1 accounts for every integer that includes two 1's and two 2's.
To account for the other cases, we must count the number of options for the digits in red.
The digits in red could be composed of any 2 of the 4 digits.
Number of ways to choose 2 digits from 4 = 4C2 = (4*3)/(2*1) = 6.
Thus, we multiply by 6:
180*6 = 1080.

Total options = 480+1080 = 1560.

The correct answer is C.

GMATGuruNY wrote:

parveen110 wrote:The total number of natural numbers of six digits that can be formed with digits 1,2,3,4, if all the digits are to appear in the same number at least once, is:
a. 740
b. 1830
c. 1560
d. none of these

Number of ways to arrange n distinct elements = n!.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Thus:
Number of ways to arrange AAABB = 5!/(3!2!).
We divide by 3! to account of the 3 identical A's and by 2! to account for the 2 identical B's.

Option A: 3 digits the same, the other 3 different
Case 1:
Let's say the 6-digit integer is to be composed of 1,1,1,2,3,4.
Number of ways to arrange the 6 digits = 6!/3! = 120.
Other cases:
The digit in red could be 1, 2, 3, or 4, for a total of 4 options.
Thus, we multiply by 4:
120*4 = 480.

Option B: 2 digits the same, another 2 digits the same, the other 2 different
Case 1:
Let's say the 6-digit integer is to be composed of 1,1,2,2,3,4.
Number of ways to arrange the 6 digits = 6!/(2!2!) = 180.
Other cases:
Case 1 accounts for every integer that includes two 1's and two 2's.
To account for the other cases, we must count the number of options for the digits in red.
The digits in red could be composed of any 2 of the 4 digits.
Number of ways to choose 2 digits from 4 = 4C2 = (4*3)/(2*1) = 6.
Thus, we multiply by 6:
180*6 = 1080.

Total options = 480+1080 = 1560.

The correct answer is C.

Thank You for posting, Mitch.
Could you please explain as to what the question is actually asking..i am not able to understand the language. I should have mentioned it while posting the question. Thanks.

parveen110 wrote: Thank You for posting, Mitch.
Could you please explain as to what the question is actually asking..i am not able to understand the language. I should have mentioned it while posting the question. Thanks.

Question rephrased:
How many 6-digit integers can be formed from the digits 1, 2, 3, and 4 if the integer must contain at least one 1, at least one 2, at least one 3, and at least one 4?