How many 5 digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.

Q: How many 5 digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.

Ans: We can choose the digit that repeats in 3c1 ways. The remaining two digits in the 5 digits number can either be same or different. If they are same, there are 2 ways of choosing the digit. In total there would be:

3c1 * ( 2* 5!/(2!*3!) + 5!/(3!)) = 120.

But the answer is 90. Can some one please explain how.

• But the two remaining digits can´t be the same if you use all the digits of 1,2 and 3. – callculus Oct 28 '18 at 13:59
• Example of a 5 digit number with one digit repeating 3 times is 21131 or 21121. In 21131 the remaining two digits 2 and 3 are different. – Hacky wacky Oct 28 '18 at 14:02
• Yes you can interpret it like this. But you maybe also can say, that every number must contain the numbers 1,2,3. – callculus Oct 28 '18 at 14:04
• 120 is correct, you are right, the given answer is wrong. – Parcly Taxel Oct 28 '18 at 14:05
• I understand your point. But here is the question verbatim: The number of – Hacky wacky Oct 28 '18 at 14:06

Let´s assume that we don´t have to use all digits 1,2 and 3 to form the number. We define $$X,Y,Z\in \{1,2,3 \}$$

Case 1: Two different groups of digits, where each group consists of one kind of number.: $$XXXYY$$

These sequence can be arranged in $$\frac{5!}{3!\cdot 2!}=\frac{120}{12}=10$$ ways.

$$X$$ and $$Y$$ can have the following combinations: $$(1,2);(2;1);(1,3);(3,1);(2,3);(3,2)$$

Thus for case 1 we have $$6\cdot 10=60$$ ways.

Case 2: Three different groups of digits where each group consists of one kind number and one group has 3 digits: $$XXXYZ$$. This sequence can be arranged in $$\frac{5!}{3!\cdot 1!\cdot 1!}=\frac{120}{6}=20$$ ways.

$$X,Y,Z$$ can have $$3$$ combinations and for case 2 there exists $$60$$ ways.

Finally we can say that $$120$$ five digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.

• The explanation for case 1 is perfectly valid. But case 2 is slightly wrong. For the group XXXYZ, we are using all the digits in which one is repeating 3 times. So the number of ways in which you can choose the repeating number is 3. Hence there can be 20 * 3 arrangements of type XXXYZ. – Hacky wacky Oct 28 '18 at 15:11
• @Hackywacky Indeed I get 120 as well. – callculus Oct 28 '18 at 15:31