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10 Feb 2021, 05:30
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (01:58) correct
55%
(02:37)
wrong
based on 117
sessions
History
Date
Time
Result
Not Attempted Yet
Number of permutation of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time, such that the digit 1 appearing somewhere to the left of 2, 3 appearing to somewhere the left of 4 and 5 somewhere to the left of 6, is (e.g. 815723946 would be one such permutation)
(A) 9*7!
(B) 8!*4!
(C) 8!
(D) 5!*4!
(E) 4!*4!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 9*7!
(B) 8!*4!
(C) 8!
(D) 5!*4!
(E) 4!*4!
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
10 Feb 2021, 07:09
Kudos
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Bunuel
Total ways will be 9!.
Half of these will have 1 to the left of 2 and other half 2 to the left of 1.
So 1 to the left of 2 =\(\frac{9!}{2}\).
Similarly half of these will have 3 to the left of 4, and half of those 5 to the left of 6 = \(\frac{9!}{2*2*2}=\frac{9*8*7!}{8}=9*7!\)
A
General Discussion
21 Dec 2024, 06:57
Kudos
Bookmarks
2.0 Experimental Advanced. Lacks access to real-time info and some Gemini features.
Here's how to solve this problem:
1. Consider the pairs:
Since the order within the pairs (1,2), (3,4), and (5,6) is fixed (1 must be before 2, 3 before 4, and 5 before 6), treat each pair as a single unit.
2. Total arrangements without restrictions:
There are 9 digits in total, so there would be 9! ways to arrange them without any restrictions.
3. Account for the fixed order within pairs:
For the pair (1,2): In any arrangement of the 9 digits, there are 2! = 2 ways to arrange 1 and 2. Only one of these arrangements has 1 to the left of 2. So, we need to divide by 2! for this pair.
Similarly, for the pair (3,4), we need to divide by 2! to account for the fixed order of 3 being to the left of 4.
And for the pair (5,6), we need to divide by 2! again.
4. Calculate the number of valid arrangements:
The total number of valid arrangements is:
9! / (2! * 2! * 2!) = 9! / 8 = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 8 = 9 * 7!
Answer:
The number of permutations is 9 * 7!. Therefore, the correct answer is (A).
Here's how to solve this problem:
1. Consider the pairs:
Since the order within the pairs (1,2), (3,4), and (5,6) is fixed (1 must be before 2, 3 before 4, and 5 before 6), treat each pair as a single unit.
2. Total arrangements without restrictions:
There are 9 digits in total, so there would be 9! ways to arrange them without any restrictions.
3. Account for the fixed order within pairs:
For the pair (1,2): In any arrangement of the 9 digits, there are 2! = 2 ways to arrange 1 and 2. Only one of these arrangements has 1 to the left of 2. So, we need to divide by 2! for this pair.
Similarly, for the pair (3,4), we need to divide by 2! to account for the fixed order of 3 being to the left of 4.
And for the pair (5,6), we need to divide by 2! again.
4. Calculate the number of valid arrangements:
The total number of valid arrangements is:
9! / (2! * 2! * 2!) = 9! / 8 = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 8 = 9 * 7!
Answer:
The number of permutations is 9 * 7!. Therefore, the correct answer is (A).
