Pascal Triangle
Trending Questions
Q.
What is formula for
Q. If nC4, nC5, nC6 are in A.P., then the value(s) of n is/are
- 7
- 14
- 8
- 16
Q. If n+1C3=2 nC2, then the value of n is
Q. The number of value(s) of r satisfying the equation 69C3r−1− 69Cr2= 69Cr2−1− 69C3r is
Q. If n+2C6n−2P2=11, then n satisfies the equation:
- n2+2n−80=0
- n2+5n−84=0
- n2+3n−108=0
- n2+n−110=0
Q. If 2nC3 nC3=11, then the value of n is
Q. The value of 19C2 is
- 171
- 342
- 19!2!
- 19!17!
Q. If (n+1)!=12×(n−1)!, then the value of n can be
- −4
- 3
- −3
- 4
Q. If nC4, nC5, nC6 are in A.P., then the value(s) of n is/are
- 7
- 14
- 8
- 16
Q. The value of n∑r=1r×r! is
- (n+1)!−1
- (n)!−1
- (n+1)!−n
- n!
Q. Let f:(0, ∞)→R be given by
f(x)=x∫1/xe−(t+1t)dtt.
Then
f(x)=x∫1/xe−(t+1t)dtt.
Then
- f(x) is monotonically increasing on [1, ∞)
- f(x) is monotonically decreasing on (0, 1)
- f(x)+f(1x)=0, for all x∈(0, ∞)
- f(2x) is an odd function of x on R
Q. If nC5= nC10, then the value of 16Cn is
- 1
- 16
- 15
- 15!
Q. If 18C2r= 18Cr+3, then the value of r is/are
- 2
- 3
- 4
- 5
Q. If 18C2r= 18Cr+3, then the value of r is/are
- 2
- 3
- 4
- 5
Q. If the number of terms in (a+b)n2+3 and (c+d)3n+4 are same, where a, b, c, d≠0, n∈N, then the number of possible value(s) of n is
- 0
- 1
- 2
- More than 2 but finite.
Q. If nCr=84, nCr−1=36 and nCr+1=126, then the value of n is
- 9
- 12
- 11
- 10
Q. If the number of terms in the expansion of (a+b)n2+3 is 7, then the number of value(s) of n, (n∈N) is
Q.
Find the value of r, if 20Cr+1=20C3r−1
1
2
4
5
Q. If n= mC2, then the value of nC2 is
- 2 m+1C2
- 3 m+1C4
- m+1C4
- 3 m+1C2
Q. If nC2+ nC3= n+1Cr, then the minimum possible value of r is
- 0
- 3
- 2
- 1
Q. If nC4= nC5, then the value of nC2 is
- 10
- 28
- 21
- 36
Q. The value of 4∑r=0 4Cr 4Cr+ 4Cr+1 is
- 12
- 3
- 13
- 2
Q. Compute 8!6!×2!
Q. If n+1C5− nC4> nC3, then the minimum value of n is
Q. The expression nCr+2 nCr−1+ nCr−2 is equal to
- n+1Cr
- n+2Cr+1
- n+2Cr
- n+1Cr+1
Q. If nC5= nC10, then the value of 16Cn is
- 1
- 15
- 16
- 15!
Q. If nPr=nPr+1 and nCr=nCr−1, then the values of n and r are
- r, 3
- 3, 2
- 4, 2
- 3, 4
Q. The number of words that can be formed by the letters of WORDS is
Q. The value of n∑r=1r nCr nCr−1 is
- (n+1)22
- n22
- n(n+1)2
- n(n−1)2
Q.
Let a relation R be defined by R = {(4, 6); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1 oR is
{(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}
{(1, 1), (4, 4), (7, 7), (3, 3)}
{(1, 5), (1, 6), (3, 6)}
None of these