Friday, 5 January 2018
Tuesday, 2 January 2018
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Interesting Math Problem
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Combinations - 10-digit numbers of 1s,2s,3s
a) How many different
10-digit numbers can be formed from the digits 1, 2, 3 where digit 3 in each
number is found exactly three times ?
b) How many of those numbers can be divided by 6 ?
b) How many of those numbers can be divided by 6 ?
a) How many different
10-digit numbers can be formed from the digits 1, 2, 3 where digit 3 in each
number is found exactly three times ? Choose the three spots for the 3's in C(10,3) ways. 10 × 9 × 8 /
3!
For the remaining 7 spots, they can be either a 1 or a 2, so there are 2^7 possibilities.
In total we have (10 × 9 × 8 / 3!) × 2^7 = 15360
b) How many of them are divisible by 6 ?
In order to be divisible by 6, the number must be even and divisible by 3.
If the only available digits are 1,2,3, then to be even, the number must
end in 2.
To be divisible by 3, the sum of the digits must also be divisible by 3.
So we start with three 3's and a 2.
The sum of the remaining six digits must have a remainder of 1 (mod 3).
For the sum, 1 and 4 are < 6 × 1 and 13 > 6 × 2, so the only
possibilities are 7 and 10.
The remaining six digits (in any order) can be:
2 2 2 2 1 1 sum of 10
2 1 1 1 1 1 sum of 7
So we combine each of these with the three 3's,
and we get
9! /(3! 4! 2!) + 9! / (3! 1! 5!) = 1260 + 504 = 1764
(The 10th digit is always a 2.)
Another way of looking at this is:
first assign the threes to C(9, 3) places,
then assign the twos to C(6, 4) or C(6, 1) places.
Then we get
9 × 8 × 7 / 3! × (6 × 5 × 4 × 3 / 4! + 6 / 1!) = 84 × 21 = 1764
For the remaining 7 spots, they can be either a 1 or a 2, so there are 2^7 possibilities.
In total we have (10 × 9 × 8 / 3!) × 2^7 = 15360
b) How many of them are divisible by 6 ?
In order to be divisible by 6, the number must be even and divisible by 3.
If the only available digits are 1,2,3, then to be even, the number must
end in 2.
To be divisible by 3, the sum of the digits must also be divisible by 3.
So we start with three 3's and a 2.
The sum of the remaining six digits must have a remainder of 1 (mod 3).
For the sum, 1 and 4 are < 6 × 1 and 13 > 6 × 2, so the only
possibilities are 7 and 10.
The remaining six digits (in any order) can be:
2 2 2 2 1 1 sum of 10
2 1 1 1 1 1 sum of 7
So we combine each of these with the three 3's,
and we get
9! /(3! 4! 2!) + 9! / (3! 1! 5!) = 1260 + 504 = 1764
(The 10th digit is always a 2.)
Another way of looking at this is:
first assign the threes to C(9, 3) places,
then assign the twos to C(6, 4) or C(6, 1) places.
Then we get
9 × 8 × 7 / 3! × (6 × 5 × 4 × 3 / 4! + 6 / 1!) = 84 × 21 = 1764
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