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If a 3-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number (which is a 3 -digit number) is divisible by 4 and 5 ?
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The correct answer is:
$\frac{29}{180}$
We need 3 -digit number which is divisible by $4 \& 5$ both.
i.e. their last two digits are
$00,20,40,60 \& 80$
Now, ending with 00 are '9'.
$\{100,200, \ldots \ldots, 900\}$
If digit repeat other than ' 0 ' then they are
$\{220,440,660,880\}$
but 220 numbers can be permuted according to the condition as $\{220,202\}$
So, there are ' 8 ' other favorable cases.
If the number have no digit repeated like 320 .
320 can be permuted in 4 ways.
$\{302,230,320,203\}$
So, such numbers are $8 \times 4 \times 4=128$
Total favorable $=9+8+128=145$
So, required prob. $=\frac{145}{900}=\frac{29}{180}$
i.e. their last two digits are
$00,20,40,60 \& 80$
Now, ending with 00 are '9'.
$\{100,200, \ldots \ldots, 900\}$
If digit repeat other than ' 0 ' then they are
$\{220,440,660,880\}$
but 220 numbers can be permuted according to the condition as $\{220,202\}$
So, there are ' 8 ' other favorable cases.
If the number have no digit repeated like 320 .
320 can be permuted in 4 ways.
$\{302,230,320,203\}$
So, such numbers are $8 \times 4 \times 4=128$
Total favorable $=9+8+128=145$
So, required prob. $=\frac{145}{900}=\frac{29}{180}$
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