How many numbers are there between \[100{\text{ and }}500\]with the digits \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5?{\text{ }}\]if,
(i) repetition of digit is allowed
(ii) the repetition of digits is not allowed
Answer
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Hint: In order to calculate the number of the numbers there between\[100{\text{ and }}500\] with the digits \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\], we must notice that we are dealing with 3 digit numbers.
Complete step by step answer:
Case (i) When repetition of digit is allowed
We are considering 3 digit numbers.
Let us solve this problem by taking cases, and the three digit spaces be represented by _ _ _
So,
\[\begin{array}{*{20}{l}}
{1{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{2{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{3{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{4{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}}
\end{array}\]
Then,
Total number of ways,
$ = 4 \times 6 \times 6 $
$ = 144 - 1 $
(Because here we took the case of 100 to which is equal to 100).
$ = 143$ ways
Case (ii) When repetition is not allowed,
Then,
\[\begin{array}{*{20}{l}}
{1{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}} \\
{2{\text{ }}\_{\text{ }}\_{\text{ = }}5 \times 4{\text{ ways}}} \\
{3{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}} \\
{4{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}}
\end{array}\]
Therefore, the total number of ways = $80$ ways
Additional Information:
A permutation is an arrangement of objects in a definite order. The members or elements of sets are arranged here in a sequence or linear order. Permutation can be classified into three different categories: Permutation of n different objects (when repetition is not allowed), Repetition, where repetition is allowed, Permutation when the objects are not distinct (Permutation of multisets)
Note:
It should be taken into consideration that we need to solve it first to find all the possible numbers between 100 and 500 with all the digits. Thereafter subtract the digits withs the repetition and we will get the solution for when repetition is not allowed. This question can be also solved with different methods using the permutation formula or as per the requirement.
Complete step by step answer:
Case (i) When repetition of digit is allowed
We are considering 3 digit numbers.
Let us solve this problem by taking cases, and the three digit spaces be represented by _ _ _
So,
\[\begin{array}{*{20}{l}}
{1{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{2{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{3{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}} \\
{4{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}6 \times 6{\text{ ways}}}
\end{array}\]
Then,
Total number of ways,
$ = 4 \times 6 \times 6 $
$ = 144 - 1 $
(Because here we took the case of 100 to which is equal to 100).
$ = 143$ ways
Case (ii) When repetition is not allowed,
Then,
\[\begin{array}{*{20}{l}}
{1{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}} \\
{2{\text{ }}\_{\text{ }}\_{\text{ = }}5 \times 4{\text{ ways}}} \\
{3{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}} \\
{4{\text{ }}\_{\text{ }}\_{\text{ }} = {\text{ }}5 \times 4{\text{ ways}}}
\end{array}\]
Therefore, the total number of ways = $80$ ways
Additional Information:
A permutation is an arrangement of objects in a definite order. The members or elements of sets are arranged here in a sequence or linear order. Permutation can be classified into three different categories: Permutation of n different objects (when repetition is not allowed), Repetition, where repetition is allowed, Permutation when the objects are not distinct (Permutation of multisets)
Note:
It should be taken into consideration that we need to solve it first to find all the possible numbers between 100 and 500 with all the digits. Thereafter subtract the digits withs the repetition and we will get the solution for when repetition is not allowed. This question can be also solved with different methods using the permutation formula or as per the requirement.
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