NIMCET 2008 — Mathematics PYQ
NIMCET | Mathematics | 2008An eight-digit number divisible by 9 is to be formed by using 8 digits out of the digits without replacement. The number of ways in which this can be done is:
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An eight-digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,…,9 without replacement. The number of ways in which this can be done is:
9!
2(7!)
4(7!)
36(7!)
(Correct Answer)36(7!)
1. The Divisibility Rule for 9
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
2. Finding the Total Sum of Available Digits
The set of available digits is {0,1,2,3,4,5,6,7,8,9}.
The sum of all these 10 digits is:
3. Selecting 8 Digits
Since we need to form an 8-digit number, we must exclude 2 digits from the set. Let the sum of the 8 selected digits be S8 and the sum of the 2 excluded digits be S2.
For S8 to be divisible by 9, S2 must also be divisible by 9 (because 45 is already divisible by 9).
4. Possible Excluded Pairs (S2=9 or 18)
We look for pairs {x,y} from the set {0,1,…,9} whose sum is 9 or 18:
Sum = 9: {0,9},{1,8},{2,7},{3,6},{4,5}
Sum = 18: {9,9} (Not possible as replacement is not allowed).
So, there are 5 possible cases for the set of 8 digits:
Exclude {0,9}: Remaining digits {1,2,3,4,5,6,7,8}
Exclude {1,8}: Remaining digits {0,2,3,4,5,6,7,9}
Exclude {2,7}: Remaining digits {0,1,3,4,5,6,8,9}
Exclude {3,6}: Remaining digits {0,1,2,4,5,7,8,9}
Exclude {4,5}: Remaining digits {0,1,2,3,6,7,8,9}
5. Calculating Arrangements for Each Case
Case 1 (No zero):
The digits are {1,2,3,4,5,6,7,8}.
Number of arrangements = 8!
Cases 2, 3, 4, and 5 (Includes zero):
In each of these 4 cases, we have 8 digits including '0'. An 8-digit number cannot start with 0.
Ways to fill the first place (non-zero): 7 ways.
Remaining 7 places: 7! ways.
Total for one case = 7×7!
Total for all 4 cases = 4×(7×7!)
6. Total Number of Ways
Correct Option: (d) 36(7!)
1. The Divisibility Rule for 9
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
2. Finding the Total Sum of Available Digits
The set of available digits is {0,1,2,3,4,5,6,7,8,9}.
The sum of all these 10 digits is:
3. Selecting 8 Digits
Since we need to form an 8-digit number, we must exclude 2 digits from the set. Let the sum of the 8 selected digits be S8 and the sum of the 2 excluded digits be S2.
For S8 to be divisible by 9, S2 must also be divisible by 9 (because 45 is already divisible by 9).
4. Possible Excluded Pairs (S2=9 or 18)
We look for pairs {x,y} from the set {0,1,…,9} whose sum is 9 or 18:
Sum = 9: {0,9},{1,8},{2,7},{3,6},{4,5}
Sum = 18: {9,9} (Not possible as replacement is not allowed).
So, there are 5 possible cases for the set of 8 digits:
Exclude {0,9}: Remaining digits {1,2,3,4,5,6,7,8}
Exclude {1,8}: Remaining digits {0,2,3,4,5,6,7,9}
Exclude {2,7}: Remaining digits {0,1,3,4,5,6,8,9}
Exclude {3,6}: Remaining digits {0,1,2,4,5,7,8,9}
Exclude {4,5}: Remaining digits {0,1,2,3,6,7,8,9}
5. Calculating Arrangements for Each Case
Case 1 (No zero):
The digits are {1,2,3,4,5,6,7,8}.
Number of arrangements = 8!
Cases 2, 3, 4, and 5 (Includes zero):
In each of these 4 cases, we have 8 digits including '0'. An 8-digit number cannot start with 0.
Ways to fill the first place (non-zero): 7 ways.
Remaining 7 places: 7! ways.
Total for one case = 7×7!
Total for all 4 cases = 4×(7×7!)
6. Total Number of Ways
Correct Option: (d) 36(7!)