A three-digit even number is to be formed from given 6 digits 1, 2, 3, 4, 6, 7 The number at one's place can be filled by 2, 4, 6. Since repetition is not allowed Now, Tens place can be filled by the remaining 5 digits So, Tens place can be filled in 5 ways Similarly, Hundred place can be filled by the remaining 4 digits So, Hundred place can be filled in 4 ways So, the Required number of ways in which three-digit even numbers can be formed from the given digits is ⇒ Number of ways = 5 × 4 × 3 ⇒ Number of ways = 60
So basically, I attempted this question as- There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place. So, the total choices are, by multiplication principle- $$4*4*4=64$$ And well and good, this was the answer. But what if I reversed the method? So I take some particular numbers, like $1,2,3$ and say that, well, $1$ can go in $3$ places, $2$ in $2$ places and $3$ in $1$ place, so by multiplication principle, there are $6$ ways of forming a $3$-digit number with $1,2,3$. But there are $4$ different numbers. So the number of $3$-number combinations are- $(1,2,3)$,$(1,2,4)$,$(1,3,4)$,$(2,3,4)$. Each can be arranged in $6$ ways, so we get $24$ ways totally. So why is my answer different here? Answer VerifiedHint: Fundamental principle of counting: According to the fundamental principle of counting if a task can be done in “m” ways and another task can be done in “n” ways, then the number of ways in which both the tasks can be done in mn ways.Complete step by step solution:Case [1]: Repetition not allowed:The number of ways in which the ones place can be filled = 5 ways.The number of ways in which the tens place can be filled = 4 ways (because repetition is not allowed. So, the choice of one place cannot be used). The number of ways in which hundreds place can be filled = 3 ways.Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 4\times 3=60$Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 60 i.e 5! waysCase [2]: Repetition is allowed:The number of ways in which the ones place can be filled = 5 ways.The number of ways in which the tens place can be filled = 5 ways (because repetition is allowed. So, the choice of one's place can be used). The number of ways in which hundreds place can be filled = 5 ways.Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 5\times 5=125$ Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 125Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = ${}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ which is the same result as above. Text Solution Solution : (i) When repetition of digits is allowed: <br> No. of ways of choosing firsy digits = 5 <br> No. of ways of choosing second digit = 5 <br> No. of ways of choosing third digit = 5 <br> Therefore, total possible numbers `= 5 xx 5 xx 5 = 125` <br> (ii) When repetition of digits is not allowed: <br> No. of ways of choosing first digit = 5 <br> No. of ways of choosing second digit = 4 <br> No. of ways of choosing thrid digit = 3 <br> Total possible numbers `= 5 xx 4 xx 3 = 60`. |