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11 Nov 2019, 02:47
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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52% (02:43) correct
48%
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based on 204
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How many positive integers less than 9999 are there in which the sum of the digits equals 6?
(A) 55
(B) 60
(C) 61
(D) 84
(E) 120
Are You Up For the Challenge: 700 Level Questions
(A) 55
(B) 60
(C) 61
(D) 84
(E) 120
Are You Up For the Challenge: 700 Level Questions
11 Nov 2019, 06:03
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The posted question is virtually the same as the following:
How many ways can 6 identical chocolates be distributed among 4 children?
We can use the SEPARATOR METHOD.
6 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 6 chocolates and 3 separators:
O|O|OO|OO
Every arrangement of the elements above represents one way to distribute the 6 chocolates among 4 children A, B, C and D:
O|O|OO|OO = A gets 1 chocolate, B gets 1 chocolate, C gets 2 chocolates. D gets 2 chocolates
This case is the equivalent of the following integer: 1122
||OO|OOOO = A gets 0 chocolates, B gets 0 chocolates, C gets 2 chocolates. D gets 4 chocolates
This case is the equivalent of the following integer: 0024 = 24
OOOOOOOOOO||| = A gets all 6 chocolates
This case is the equivalent of the following integer: 6000
To count the number of possible distributions, we need to calculate the number of ways to position the 3 separators.
Since there are 6 chocolates and 3 separators -- yielding a total of 9 positions -- the 3 separators must occupy 3 of the 9 positions.
From 9 positions, the number of ways to choose 3 to assign to the 3 separators = 9C3 \(= \frac{9*8*7}{3*2*1} = 84 \)
How many ways can 6 identical chocolates be distributed among 4 children?
We can use the SEPARATOR METHOD.
6 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 6 chocolates and 3 separators:
O|O|OO|OO
Every arrangement of the elements above represents one way to distribute the 6 chocolates among 4 children A, B, C and D:
O|O|OO|OO = A gets 1 chocolate, B gets 1 chocolate, C gets 2 chocolates. D gets 2 chocolates
This case is the equivalent of the following integer: 1122
||OO|OOOO = A gets 0 chocolates, B gets 0 chocolates, C gets 2 chocolates. D gets 4 chocolates
This case is the equivalent of the following integer: 0024 = 24
OOOOOOOOOO||| = A gets all 6 chocolates
This case is the equivalent of the following integer: 6000
To count the number of possible distributions, we need to calculate the number of ways to position the 3 separators.
Since there are 6 chocolates and 3 separators -- yielding a total of 9 positions -- the 3 separators must occupy 3 of the 9 positions.
From 9 positions, the number of ways to choose 3 to assign to the 3 separators = 9C3 \(= \frac{9*8*7}{3*2*1} = 84 \)
11 Nov 2019, 05:47
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I will break 6 into 4 digits first then find the number of satisfactory numbers for each set of numbers.
To break 6 down, we can have:
(1) 6 + 0 + 0 + 0. Note 0006 also counts, so this is 4 numbers. We can use 4! / (3!*1!) = 4 for all of these. (3! is to get rid of the duplicate 0's.)
(2) 5 + 1 + 0 + 0. 4!/(2!) = 12.
(3) 4 + 2 + 0 + 0. Similar as above, 12 numbers.
(4) 4 + 1 + 1 + 0. 4! / 2! = 12 numbers.
(5) 3 + 3 + 0 + 0. 4! / (2! * 2!) = 6.
(6) 3 + 2 + 1 + 0. 4! = 24.
(7) 3 + 1 + 1 + 1. 4! / 3! = 4.
(8) 2 + 2 + 2 + 0. 4! / 3! = 4.
(9) 2 + 2 + 1 + 1. 4! / (2!*2!) = 6.
In total, we have 4 + 36 + 6 + 24 + 8 + 6 = 40 + 30 + 14 = 84 numbers.
Another quick way to do this is to imagine a number that starts with 0000 and we are adding 1's in front of the slots to make the digits sum to 6, and we add 1's 6 times. For example 110, 110, 110, 0. Would correspond to 2220. 0, 0, 1110, 1110 would correspond to 0033. Regardless we can add 1's everywhere, so the first 9 digits (not including the very last 0 since we only add 1's in front of 0's) can be randomized. There are 6 one's and 3 zero's however so in total there are 9! / (6!*3!) = 9C3 = 9 * 8 * 7 / 6 = 3 * 4 * 7 = 84 combinations that are unique.
Ans: D
To break 6 down, we can have:
(1) 6 + 0 + 0 + 0. Note 0006 also counts, so this is 4 numbers. We can use 4! / (3!*1!) = 4 for all of these. (3! is to get rid of the duplicate 0's.)
(2) 5 + 1 + 0 + 0. 4!/(2!) = 12.
(3) 4 + 2 + 0 + 0. Similar as above, 12 numbers.
(4) 4 + 1 + 1 + 0. 4! / 2! = 12 numbers.
(5) 3 + 3 + 0 + 0. 4! / (2! * 2!) = 6.
(6) 3 + 2 + 1 + 0. 4! = 24.
(7) 3 + 1 + 1 + 1. 4! / 3! = 4.
(8) 2 + 2 + 2 + 0. 4! / 3! = 4.
(9) 2 + 2 + 1 + 1. 4! / (2!*2!) = 6.
In total, we have 4 + 36 + 6 + 24 + 8 + 6 = 40 + 30 + 14 = 84 numbers.
Another quick way to do this is to imagine a number that starts with 0000 and we are adding 1's in front of the slots to make the digits sum to 6, and we add 1's 6 times. For example 110, 110, 110, 0. Would correspond to 2220. 0, 0, 1110, 1110 would correspond to 0033. Regardless we can add 1's everywhere, so the first 9 digits (not including the very last 0 since we only add 1's in front of 0's) can be randomized. There are 6 one's and 3 zero's however so in total there are 9! / (6!*3!) = 9C3 = 9 * 8 * 7 / 6 = 3 * 4 * 7 = 84 combinations that are unique.
Ans: D
