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.

  • $\begingroup$ But the two remaining digits can´t be the same if you use all the digits of 1,2 and 3. $\endgroup$ – callculus Oct 28 '18 at 13:59
  • $\begingroup$ 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. $\endgroup$ – Hacky wacky Oct 28 '18 at 14:02
  • $\begingroup$ Yes you can interpret it like this. But you maybe also can say, that every number must contain the numbers 1,2,3. $\endgroup$ – callculus Oct 28 '18 at 14:04
  • 1
    $\begingroup$ 120 is correct, you are right, the given answer is wrong. $\endgroup$ – Parcly Taxel Oct 28 '18 at 14:05
  • $\begingroup$ I understand your point. But here is the question verbatim: The number of $\endgroup$ – 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.

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  • $\begingroup$ 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. $\endgroup$ – Hacky wacky Oct 28 '18 at 15:11
  • $\begingroup$ @Hackywacky Indeed I get 120 as well. $\endgroup$ – callculus Oct 28 '18 at 15:31

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