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My attempt : Fixing $2$ and $3$ at the tens and units place. Hundreds place can be filled up by $(1,4)$. Hence $2$ numbers can be formed in this case. Fixing $2$ and $3$ at the hundreds and tens place. Units place can be filled up by $(0,1,4)$. Hence $3$ numbers can be formed in this case. Fixing $2$ at the hundreds place and $3$ at the units place. Tens place can be filled up by $(0,1,4)$. Hence $3$ numbers can be formed. Now we,ll swap the positions of $3$ and $2$ and again apply the same method. Hence there will be a total of $3+3+2+3+3+2 = 16$ numbers that can be formed. But if I apply the formula ${}^{n-k}\mathrm P_{r-k}\times{}^r\mathrm P_k$ which is used to find the number of permutations of $n$ dissimilar things taken $r$ at a time when $k$ particular things always occur, the answer comes out to be $18$. Is it because the results $023$ and $032$ are also included which need to be subtracted as $0$ can't be the leading digit? Is my method right?

If I use the formula $$s![r-(s-1)]\cdot (^{n-s}P_{r-s})$$ which is also used to find the number of permutations of n different objects taken r at a time when s particular things are to be always included in each arrangement, the answer comes out to be 12. Whats the cause of difference?

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Yes, your method is correct, and yes, the discrepancy between the two answers is simply that the formula counts $023$ and $032$, which are not $3$-digit numbers.

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  • $\begingroup$ If I use the fomula $$ s![r-(s-1)]\cdot (^{n-s}P_{r-s}) $$ which is used to find the number of permutations of n different objects taken r at a time when s particular things are to be always included in each arrangement, the answer comes out to be 12. What is the cause of difference ? $\endgroup$ – John Jul 2 '17 at 11:48
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Consider this method.

For our third digit, we have $0$, $1$, or $4$.

Each choice creates $6$ possible numbers.

So we have $(3)(6)=18$ numbers.

Except $023$ and $032$. Not $3$ digit numbers.

$18-2=\boxed{16}$ numbers.


So yes, your method is correct.

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