Question

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Plz answer this Q!!

1*) The number of 6 digits that can be formed using the digits 0 , 1 , 2 , 5 , 7 and 9 which are divisible by 11 and no digits is repeated , is ?

(A) 60
(B) 72
(C) 48
(D) 36

Plz answer with full explanation..............
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Answer 1

Answer:

Option ( A ) 60

Step-by-step explanation:

We Have :-

Digits 0 , 1 , 2 , 5 , 7 and 9

Number formed from these digits is divisible by 11

To Find :-

Number of digits

Solution :-

For a Number ABCDEF to be divisible by 11

The divisibility rule is

| ( A + C + E ) - ( B + D + F ) | should be a multiple of 11

Sum of the digits given to us :-

0 + 1 + 2 + 5 + 7 + 9

24

We have 6 digits

A + C + E = B + D + F = 12

1st Way :-

{ 0 , 5 , 7 } = { A , C , E }

{ 1 , 2 , 9 } = { B , D , F }

2 × 2! × 3! = 24

2nd Way :-

{ 0 , 5 , 7 } = { B , D , F }

{ 1 , 2 , 9 } = { A , C , E }

3! × 3! = 36

So total Number of Ways :-

24 + 36

60

So Option ( A ) 60 is the correct option

Answer 2

Answer:

question ❓❓❓⬇️

1*) The number of 6 digits that can be formed using the digits 0 , 1 , 2 , 5 , 7 and 9 which are divisible by 11 and no digits is repeated , is ?

Step-by-step explanation:

answer..⬇️

b..60..

Hint: Find the possible values of the numbers that odd places and even places can take. Make all the possible cases. Calculate the total number of ways for each case, then add the results of both the cases to get the required answer.

solve with explanation..

First of all, write the divisibility rule of 11.

Let the digit is written as abcdefabcdef, then, the digit is divisible by 11 if and if (a+c+e)−(b+d+f)(a+c+e)−(b+d+f) is 0 or a multiple of 11.

We have possible that a,b,c,d,e,fa,b,c,d,e,f can take to satisfy the divisibility of 11.

If each of a,c,ea,c,e can take any value from 7,5,0. Then b,d,fb,d,f can take 9,2 or 1.

But, aa cannot take 0 as it will make a five-digit number then.

aa has 2 options, cc has 2 options, if any of them is taken by aa and ee will be left with only 1 option.

Similarly, bb can take 3 options, dd has 2 options and ff has one option left.

Total cases are for abcdefabcdef when a,c,ea,c,e can take any value from 7,5,0 and b,d,fb,d,f can take 9,2 or 1are 2×3×2×2×1×1=242×3×2×2×1×1=24

Similarly take the case when b,d,fb,d,f can take any value from 7,5,0 and a,c,ea,c,e can take 9,2 or1.

bb can take 3 options, dd has 2 options and ff has one option left

And similarly aa can take 3 options, cc has 2 options and ee has one option left.

Hence, total number of ways for this case is 3×3×2×2×1×1=363×3×2×2×1×1=36

Therefore, the number of 6 digit 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 sum of 24 and 36, which is 60.

Hence, option 2 is correct.

correct answer ⬇️

b.60....

note

11. The divisibility rule of 11 states that the difference of sum of odd places and sum of even places must be a multiple of 11 or should be 0.

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