Find the sum of four-digit numbers that can be formed using the digits 0,2,4,7,8 without repetition.
Answer
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Hint: The question calls for the answer to be without repetition, so we need to solve the sum in that way always reducing the numbers as we go on multiplying.
Complete step-by-step answer:
The number of $4-$ digit numbers formed by using 0,2,4,7,8 without repetition ${ }^{5} P_{4}-{ }^{4} P_{3}=120-24=96$
Out of these 96 numbers,
$=545958$ numbers contain 2 in units place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in tens place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in hundreds place.
${ }^{4} P$ numbers contain 2 in units place.
$\therefore$ The values obtained by adding 2 in all numbers.
$$
\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 2+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 20+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 200+{ }^{4} P_{2} \times 2000
$$
$={ }^{4} P_{3}(2+20+200+2000)-{ }^{3} P_{2}(2+20+200)$
$=24 \times 2222-6 \times 222$
$=24 \times 2 \times 1111-6 \times 2 \times 111$
Similarly, the value obtained by adding 4 is $24 \times 4 \times 1111-6 \times 4 \times 111$.
The value obtained by adding 7 is $24 \times 7 \times 1111-6 \times 7 \times 111$
The value obtained by adding 8 is $24 \times 8 \times 1111-6 \times 8 \times 111$
Therefore,
The sum of all numbers
$=(24 \times 2 \times 1111-6 \times 2 \times 111)+(24 \times 4 \times 1111-6 \times 4 \times 111)+(24 \times 7 \times 1111-6 \times 7 \times 111)+(24 \times 8 \times 1111-6 \times 8 \times 111)$
$=24 \times 1111 \times(2+4+7+8)-6 \times 111 \times(2+4+7+8)$
$=26664 \times 21-666 \times 21$
$=559944-13986$
$=545958$
Therefore, this is the answer after solving the sum using permutation and combination formulas.
Note: A permutation of an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Before, solving the sum a student needs to understand the meaning of the word permutation and how to solve them.
Complete step-by-step answer:
The number of $4-$ digit numbers formed by using 0,2,4,7,8 without repetition ${ }^{5} P_{4}-{ }^{4} P_{3}=120-24=96$
Out of these 96 numbers,
$=545958$ numbers contain 2 in units place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in tens place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in hundreds place.
${ }^{4} P$ numbers contain 2 in units place.
$\therefore$ The values obtained by adding 2 in all numbers.
$$
\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 2+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 20+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 200+{ }^{4} P_{2} \times 2000
$$
$={ }^{4} P_{3}(2+20+200+2000)-{ }^{3} P_{2}(2+20+200)$
$=24 \times 2222-6 \times 222$
$=24 \times 2 \times 1111-6 \times 2 \times 111$
Similarly, the value obtained by adding 4 is $24 \times 4 \times 1111-6 \times 4 \times 111$.
The value obtained by adding 7 is $24 \times 7 \times 1111-6 \times 7 \times 111$
The value obtained by adding 8 is $24 \times 8 \times 1111-6 \times 8 \times 111$
Therefore,
The sum of all numbers
$=(24 \times 2 \times 1111-6 \times 2 \times 111)+(24 \times 4 \times 1111-6 \times 4 \times 111)+(24 \times 7 \times 1111-6 \times 7 \times 111)+(24 \times 8 \times 1111-6 \times 8 \times 111)$
$=24 \times 1111 \times(2+4+7+8)-6 \times 111 \times(2+4+7+8)$
$=26664 \times 21-666 \times 21$
$=559944-13986$
$=545958$
Therefore, this is the answer after solving the sum using permutation and combination formulas.
Note: A permutation of an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Before, solving the sum a student needs to understand the meaning of the word permutation and how to solve them.
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