
You have the following digits 4, 5, 6, 0, 7 and 8. Using them, make five numbers each with 6 digits.
Answer
522.6k+ views
Hint: We have 6 digits. We can make 6 digits numbers by giving each of the digits to each place in the 6-digit number.
Complete step by step solution: We have the digits 4, 5, 6, 0, 7 and 8. We can make 6-digit numbers by placing the given digits in each of the places of the six-digit number. We cannot put 0 in the ${1}^{\text{st}}$ place as it will make the number a 5-digit number.
We can write five numbers each with 6 digits as follows;
456078
560784
607845
784560
845607
Additional information:
The number of 6 digits numbers that can be formed using the given digits is given by,
\[{5 \times 5 \times 4 \times 3 \times 2 \times 1}\]
The ${1}^{\text{st}}$ digit can be any of the digits except 0. The 2nd place can have any of the other 5 digits not used in the ${1}^{\text{st}}$ place. Similarly, the ${3}^{\text{st}}$, ${4}^{\text{st}}$, ${5}^{\text{st}}$ and ${6}^{\text{st}}$ place have any 4, 3, 2 and 1 of the given digits respectively. So, the total number of 5-digit numbers that can be formed using the given digits is,
\[{5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600}\]
Note: The answer need not be unique. We can form any 5 digits number by randomly placing the given digits. We must make sure that 0 does not come in the ${1}^{\text{st}}$ place. We use the concept of permutations to find the number of 6-digit numbers that can be formed using the given digits.
Complete step by step solution: We have the digits 4, 5, 6, 0, 7 and 8. We can make 6-digit numbers by placing the given digits in each of the places of the six-digit number. We cannot put 0 in the ${1}^{\text{st}}$ place as it will make the number a 5-digit number.
We can write five numbers each with 6 digits as follows;
456078
560784
607845
784560
845607
Additional information:
The number of 6 digits numbers that can be formed using the given digits is given by,
\[{5 \times 5 \times 4 \times 3 \times 2 \times 1}\]
The ${1}^{\text{st}}$ digit can be any of the digits except 0. The 2nd place can have any of the other 5 digits not used in the ${1}^{\text{st}}$ place. Similarly, the ${3}^{\text{st}}$, ${4}^{\text{st}}$, ${5}^{\text{st}}$ and ${6}^{\text{st}}$ place have any 4, 3, 2 and 1 of the given digits respectively. So, the total number of 5-digit numbers that can be formed using the given digits is,
\[{5 \times 5 \times 4 \times 3 \times 2 \times 1 = 600}\]
Note: The answer need not be unique. We can form any 5 digits number by randomly placing the given digits. We must make sure that 0 does not come in the ${1}^{\text{st}}$ place. We use the concept of permutations to find the number of 6-digit numbers that can be formed using the given digits.
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