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27 Jun 2024, 22:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (02:55) correct
49%
(03:12)
wrong
based on 84
sessions
History
Date
Time
Result
Not Attempted Yet
How many 6 digit numbers can be formed from digits 5, 6, 7, 8 such that each digit is used at least once?
A. 792
B. 1560
C. 1650
D. 1001
E. 800
A. 792
B. 1560
C. 1650
D. 1001
E. 800
28 Jun 2024, 00:51
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Aditi_03
Since each digit is used at least once, we'd have 5 - 6 - 7 - 8 - ? - ?. We'd have two cases:
- The two extra letters (?) are equal: 55, 66, 77, or 88 (for example, 5 - 6 - 7 - 8 - 5 - 5). The number of permutations in such cases would be 6!/3! = 120. Thus, for this case, we'd have 4*120 = 480.
- The two extra letters (?) are NOT equal (for example, 5 - 6 - 7 - 8 - 5 - 6). The number of ways to select two different numbers from 4 is given by 4C2 = 6. The number of permutations in such cases would be 6!/(2!2!) = 180. Thus, for this case, we'd have 6*180 = 1,080.
Total = 480 + 1,080 = 1,560.
Answer: B.
General Discussion
28 Jun 2024, 00:07
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Answer is option B. 1560
here is the explanation
1. Total number of 6-digit combinations using 4 digits:
Each of the 6 positions in the number can be filled by any of the 4 digits (5, 6, 7, 8). Therefore, the total number of 6-digit numbers is:
4^6 = 4096
2.Subtract the cases where at least one digit is missing:
If one digit is missing, there are 3 choices for the remaining digits, and each of the 6 positions can be filled by any of these 3 digits:
3^6 = 729
There are 4 ways to choose which digit is missing, so the number of cases where exactly one digit is missing is:
4 × 729 = 2916
3. Add the cases where exactly two digits are missing (because they were subtracted twice):
If two digits are missing, there are 2 choices for the remaining digits, and each of the 6 positions can be filled by either of these 2 digits:
2^6 = 64
There are 4C2 = 6 ways to choose which 2 digits are missing, so the number of cases where exactly two digits are missing is:
6×64=384
4. Subtract the cases where exactly three digits are missing (because they were added back in three times):
If three digits are missing, there is only 1 choice for the remaining digit, and each of the 6 positions must be filled by this digit:
There are 4C3 = 4 ways to choose which 3 digits are missing, so the number of cases where exactly three digits are missing is:
4×1=4
5. Combine using the principle of inclusion-exclusion:
4096−2916+384−4=1560
which is option B
here is the explanation
1. Total number of 6-digit combinations using 4 digits:
Each of the 6 positions in the number can be filled by any of the 4 digits (5, 6, 7, 8). Therefore, the total number of 6-digit numbers is:
4^6 = 4096
2.Subtract the cases where at least one digit is missing:
If one digit is missing, there are 3 choices for the remaining digits, and each of the 6 positions can be filled by any of these 3 digits:
3^6 = 729
There are 4 ways to choose which digit is missing, so the number of cases where exactly one digit is missing is:
4 × 729 = 2916
3. Add the cases where exactly two digits are missing (because they were subtracted twice):
If two digits are missing, there are 2 choices for the remaining digits, and each of the 6 positions can be filled by either of these 2 digits:
2^6 = 64
There are 4C2 = 6 ways to choose which 2 digits are missing, so the number of cases where exactly two digits are missing is:
6×64=384
4. Subtract the cases where exactly three digits are missing (because they were added back in three times):
If three digits are missing, there is only 1 choice for the remaining digit, and each of the 6 positions must be filled by this digit:
There are 4C3 = 4 ways to choose which 3 digits are missing, so the number of cases where exactly three digits are missing is:
4×1=4
5. Combine using the principle of inclusion-exclusion:
4096−2916+384−4=1560
which is option B
