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2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these?
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2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, ...
We know ⁿpr /ⁿcr =r!
r!=336/56
r!=6
r!=3!
r=3
ⁿpr =336
n(n-1)(n-2)=336
n=8 (by verification)
r!=336/56
r!=6
r!=3!
r=3
ⁿpr =336
n(n-1)(n-2)=336
n=8 (by verification)
Community Answer
2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, ...
Given information:
nPr = 336
nCr = 56
Solution:
Understanding nPr and nCr:
- nPr (Permutations) refers to the number of ways to choose and arrange r objects from a set of n objects.
- nCr (Combinations) refers to the number of ways to choose r objects from a set of n objects without considering the order.
Relationship between nPr and nCr:
- nCr = nPr / r!
Calculating nPr and nCr:
- Given nPr = 336, we have n! / (n-r)! = 336
- Given nCr = 56, we have n! / (r! * (n-r)!) = 56
Finding n and r:
- By substituting the values of nPr and nCr, we get two equations:
1. n! / (n-r)! = 336
2. n! / (r! * (n-r)!) = 56
- By solving these two equations simultaneously, we find that n = 8 and r = 3.
Conclusion:
- Therefore, the values of n and r that satisfy the given conditions are (8, 3). Hence, the correct answer is (b) (8, 3).
nPr = 336
nCr = 56
Solution:
Understanding nPr and nCr:
- nPr (Permutations) refers to the number of ways to choose and arrange r objects from a set of n objects.
- nCr (Combinations) refers to the number of ways to choose r objects from a set of n objects without considering the order.
Relationship between nPr and nCr:
- nCr = nPr / r!
Calculating nPr and nCr:
- Given nPr = 336, we have n! / (n-r)! = 336
- Given nCr = 56, we have n! / (r! * (n-r)!) = 56
Finding n and r:
- By substituting the values of nPr and nCr, we get two equations:
1. n! / (n-r)! = 336
2. n! / (r! * (n-r)!) = 56
- By solving these two equations simultaneously, we find that n = 8 and r = 3.
Conclusion:
- Therefore, the values of n and r that satisfy the given conditions are (8, 3). Hence, the correct answer is (b) (8, 3).
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2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these?
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2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about 2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these?.
2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about 2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2. If npr = 336 and nCr = 56, then n and r will be (a) (3, 2) (b) (8, 3) (c) (7, 4) (d) none of these?.
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