MHT CET · Maths · Permutation Combination
A five digit number divisible by 3 is to be formed using the digits \(0,1,2,3,4,5\) without repetition, then the total number of ways this can be done is
- A 216
- B 240
- C 96
- D 120
Answer & Solution
Correct Answer
(A) 216
Step-by-step Solution
Detailed explanation
We know that a five digit number is divisible by 3 , if and only if sum of its digits \((=15)\) is divisible by 3. Therefore, we should not use 0 and 3 in a same number while forming the five digit numbers.
Now,
i. In case we do not use 0 , the five digit number can be formed (using digits \(1,2,3,4,5\) ) in \({ }^5 \mathrm{P}_5=120\) ways.
ii. In case we do not use 3, the five digit number can be formed (using digits \(0,1,2,4,5\) ) in \({ }^5 \mathrm{P}_5-{ }^4 \mathrm{P}_4=5!-4!=120-24=96\) ways
\(\therefore\) The total number of such 5 digit number \(=120+96=216\)
Now,
i. In case we do not use 0 , the five digit number can be formed (using digits \(1,2,3,4,5\) ) in \({ }^5 \mathrm{P}_5=120\) ways.
ii. In case we do not use 3, the five digit number can be formed (using digits \(0,1,2,4,5\) ) in \({ }^5 \mathrm{P}_5-{ }^4 \mathrm{P}_4=5!-4!=120-24=96\) ways
\(\therefore\) The total number of such 5 digit number \(=120+96=216\)
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