║⊕QUESTION⊕║
Everything around you is numbers.
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CLASS 11
PERMUTATIONS AND COMBINATIONS
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If ⁿPₓ = 6 x ⁿCₓ, find n and r when ⁿCₓ = 56
(x is actually r)
Answers
Answered by
1
Step-by-step explanation:
We know that,
nPr= n!/(n-r)!
nCr= n!/r!(n-r)!
A/Q,
nPr= 6× nCr
n!/(n-r)!=6×n!/r!(n-r)!
r!=6
r=3 (Since 3!=6)
nCr=n!/r!(n-r)!
56=n!/6(n-r)!
56=n!/6(n-3)!
56×6=n(n-1)(n-2)(n-3)!/(n-3)!
336=n{n^2-3n+2}
336=n^3-3n^2+2n
n^3-3n^2+2n-336=0
(n-8)(n^2+5n+42)=0
On solving you will get n=8
Therefore value of n=8 and value r=3
Answered by
2
Answer:
Let f:Z↦Z be defined as f(x)=x
2
,x∈Z.
We know that the square of an integer is always a unique integer.
So, ''f'' is a function.
Now, since f(−2)=f(2)=4, ''f'' is not an injection.
There is no integer x∈Z:f(x)=−1.
Hence , ''f'' is not a surjection.
Since ''f'' is neither one-one nor onto, ''f'' is not a bijection.
Step-by-step explanation:
ok
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