The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2 and 3.
A.
64322
B.
48522
C.
38664
D.
1000
Answer :
38664
Solution :
The number of numbers with 0 in the units place $$= 3! = 6.$$
The number of numbers with 1 or 2 or 3 in the units place $$= 3! - 2! = 4.$$
$$\therefore $$ The sum of the digits in the unit place $$ = 6 \times 0 + 4 \times 1 + 4 \times 2 + 4 \times 3 = 24.$$
Similarly for the tens and the hundreds places. The number of numbers with 1 or 2 or 3 in the thousands place $$= 3! .$$
$$\therefore $$ The sum of the digits in the thousands place $$ = 6 \times 1 + 6 \times 2 + 6 \times 3 = 36.$$
$$\therefore $$ The required sum $$ = 36 \times 1000 + 24 \times 100 + 24 \times 10 + 24 = 38664.$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is