ncr + nc(r-1) = n + 1cr. Prove this point.
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Answers
Answered by
2
Answer:
=
(n−r)!r!
n!
+
(n−r+1)!(r−1)!
n!
=n![
(n−r)!r!
1
+
(n−r+1)!(r−1)!
1
]
=n![
(n−r)!r(r−1)!
1
+
(n−r+1)!(r−1)!
1
]
=
(r−1)!
n!
[
(n−r)!r
1
+
(n−r+1)!
1
]
=
(r−1)!
n!
[
r(n−r)!
1
+
(n−r+1)(n−r)!
1
]
=
(r−1)!(n−r)!
n!
[
r
1
+
(n−r+1)
1
]
=
(r−1)!(n−r)!
n!
[
r(n−r+1)
n−r+1+r
]
=
(r−1)!(n−r)!
n!
[
r(n−r+1)
n+1
]
=
(n−r+1)(r−1)!r(n−r)!
(n+1)n!
=
(n−r+1)!r!
(n+1)!
=
n+1
C
r
Answered by
4
Answer:
ncr+nc(r-1) = n+1cr
125+577(r-1)=702n+1cr=802
And add it 802+702+1+1 =1506 will be the answer
Step-by-step explanation:
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