First slide

Permutations

Question

The sum of all the numbers that can be formed by writing all the digits 3, 2, 3, 4 only once is

Moderate

A

39996
B

49996
C

57776
D

None of these

Solution

The number of numbers having 2 in unit’s place $= \frac{3!}{2!}$
(because the other three places are to be filled by 3, 3, 4 of which two are identical).
The number of numbers having 3 in unit’s place = 3!, (because the other three places are to be filled by 2, 3, 4).
The number of numbers having 4 in unit’s place $= \frac{3!}{2!}$ , (because the other three places are to be filled by 3, 2, 3 of which two are identical)
∴ Sum of the digits in unit’s place
$= \frac{3!}{2!} \times 2 + 3! \times 3 + \frac{3!}{2!} \times 4 = 6 + 18 + 12 = 36$ .
Similarly, the sum of the digits in other places is 36 each.
∴ Required sum = 36 × 1000 + 36 × 100 + 36 × 10 + 36 × 1
= 36 (1000 + 100 + 10 +1) = 36 × 1111 = 39996.

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