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In each of the following 6-digit natural numbers: $333333,225522,118818,707099$, every digit in the number appears at least twice. Find the number of such 6-digit natural numbers.

This is how I'm intending to do.

1) Find the total number of 6-digit combinations.

2) Subtract the number of times there's 0 and 1 repetition of digits so that I can get at least 2 repetitions.

Total Combinations = $9 \times 10^5$

Repeat 0 Times = $9 \times 9 \times 8 \times 7 \times 6 \times 5$

Repeat 1 Time = Not sure how to calculate

The answer is 11754 but I'm struggling to get it!

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  • $\begingroup$ For 'repeat $0$ times, you seem to be counting cases with no repeated digits. I'm thinking for 'repeat $1$ time' you are calculating cases with one repeated digit (e.g., 353217). But there are other cases to consider such as two repeated digits and two singles; or a triple digit, a double digit, and a single; etc. $\endgroup$ – paw88789 Sep 20 '14 at 9:44
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First let us also accept numbers starting with a $0$.

Let $d$ denote the number of distinct digits in the number. For $d>3$ there are $0$ possibilities.

For $d=1$ there are $10$ choices of the digit and every choice leads to $1$ possibility.

For $d=3$ there are $\binom{10}{3}=120$ choices for the digits and each choice leads to $\frac{6!}{2!2!2!}=90$ possibilities.

For $d=2$ we have two split-ups.

One of them is $6=3+3$ with $\binom{10}{2}=45$ choices for the digits and each choice leads to $\binom{6}{3}=20$ possibilities.

The other is $6=4+2$. Here the chosen digits are distinghuisable. One of them is used $4$ times and the other $2$ times. So we have $10\times9=90$ choices and each choice leads to $\binom{6}{2}=\binom{6}{4}=15$ possibilities.

Adding up we find $10\times 1+120\times90+45\times20+90\times15=13060$ possibilities.

Subtracting the numbers that start with a $0$ we find $\frac{9}{10}\times13060=11754$ possibilities.

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  • $\begingroup$ Thanks drhab! I'm slightly unclear why the numbers starting with a 0 is one tenth of the total. $\endgroup$ – Faizal Ismaeel Sep 20 '14 at 10:42
  • $\begingroup$ Thought about it again and I got it! =) $\endgroup$ – Faizal Ismaeel Sep 20 '14 at 10:45
  • $\begingroup$ For d=2, for the case where one digit is used 4 times and the other 2 times, why do we have $10 \times 9$ choices? $\endgroup$ – Faizal Ismaeel Sep 29 '14 at 18:53
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    $\begingroup$ First choose the digit that appears $4$ times in the number. There are $10$ options. Then choose the digit that appears $2$ times. $9$ options are left. This gives $10\times 9$ possible choices. $\endgroup$ – drhab Sep 29 '14 at 18:59
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    $\begingroup$ No, if $d=2$ and the split up is $6=4+2$ then if the first choice is a $0$ and the second a $1$ then you have $4$ times a $0$ and $2$ times a $1$. It the first choice is a $1$ and the second a $0$ then you have $2$ times a $0$ and $4$ times a $1$. These situtations are distinct. That's why I used the term distinguishable in my answer. $\endgroup$ – drhab Sep 29 '14 at 21:15
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I think I'd suggest direct counting, with cases being: (1) same digit six times; (2) a quadruple digit and a double digit; (3) two triple digits; (4) three double digits. Things get a bit messier because if one of your digits is $0$, you need to keep it out of the leading position.

Let's try counting case (3), and I'll leave the other cases to you:

Case 3a: Two triple digits, neither being $0$: Choose two nonzero digits ($9 \choose{2}$ ways). Pick three of six positions for one of the digits ($6\choose 3$ ways). This gives a total of $36\cdot 20=720$ possibilities for this subcase.

Case 3b: Two triple digits, one being $0$: Choose a nonzero digit for the other digit (9 ways). Choose $3$ of $5$ positions for the $0$s (you have to avoid the first position). This gives $9\cdot {5\choose 3}=90$ possibilities for this subcase.

So your total for case 3 is $720+90=810$.

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  • $\begingroup$ Thanks, but I found the above answer easier to understand. =) $\endgroup$ – Faizal Ismaeel Sep 20 '14 at 10:43

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