Permutations and Combinations 3 Question 1
1. The number of 6 digits numbers that can be formed using the digits $0,1,2,5,7$ and 9 which are divisible by 11 and no digit is repeated, is
(2019 Main, 10 April I)
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Answer:
Correct Answer: 1. (a)
Solution:
- Key Idea Use divisibility test of 11 and consider different situation according to given condition.
Since, the sum of given digits
$$ 0+1+2+5+7+9=24 $$
Let the six-digit number be abcdef and to be divisible by 11 , so the difference of sum of odd placed digits and sum of even placed digits should be either 0 or a multiple of 11 means $|(a+c+e)-(b+d+f)|$ should be either 0 or $a$ multiple of 11 .
Hence, possible case is $a+c+e=12=b+d+f$ (only) Now, Case I
set ${a, c, e}={0,5,7}$ and $\operatorname{set}{b, d, f}={1,2,9}$
So, number of 6 -digits numbers $=(2 \times 2 !) \times(3 !)=24$
$[\because a$ can be selected in ways only either 5 or 7$]$.
Case II
Set ${a, c, e}={1,2,9}$ and set ${b, d, f}={0,5,7}$
So, number of 6 -digits numbers $=3 ! \times 3 !=36$
So, total number of 6 -digits numbers $=24+36=60$
