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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 13 - MCQExams.com

The value of 50C4+6r=156rC3 is?
  • 55C4
  • 55C3
  • 56C3
  • 56C4
If 32C2n1=32Cn3 then n=
  • 10
  • 9
  • 12
  • 11
The value of C20+3C21+5C22+..... to n+1 terms is
  • n2nCn
  • (2n+2)2nCn
  • (n+1)2nCn1
  • (2n+1)2n1Cn1
The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less Then 2 marks to any question is-
  • 21C7
  • 21C8
  • 21C9
  • 21C10
If (1+x)15=C0+C1xC2x2+...+C15x15, then 15C2015C21+15C2215C32+...15C215 is equal to
  • 0
  • 1
  • -1
  • none of these
The number of ways dividing 52 cards amongst four players equally, are
  • 52!(13!)4
  • 52!(13!)44!
  • 52!(12!)44!
  • None of these
If nC1=nC2, then n is equal to 
  • 2
  • 3
  • 5
  • none of these
C0+(C0+C1)+(C0+C1+C2)+......+(C0+C1+........+Cn)=
  • n.2n1
  • (n+2).2n
  • 2n
  • (n+2).2n1
If nCr=n !r !(nr) ! then the sum of the series 1+nC1+n+1C2+n+2C3+.....+n+r1Cr is equal to
  • n+rCr
  • n+r1Cr
  • n+rCr1
  • None of these
If the second term of the expansion [a1/13+aa1]n is 14a5/2 then the value of nC3nC2 is
  • 4
  • 3
  • 12
  • 6
If c02c13+c24.....+(1)ncnn+2=12013×2014 then n =
  • 2013
  • 2014
  • 2012
  • 2011
If Cr stands for nCr  then the sum of the series 2(n2)!(n2)!n![C202C21+3C22.....+(1)n(n+1)C2n],
where n is an even positive integer, is 
  • (1)n/2(n+2)
  • (1)n(n+2)
  • (1)n/2(n+1)
  • None of these
nr=0r(nr)(nCr)2 is equal to 
  • n2.2n1Cn
  • n2.2nCn1
  • 2n1Cn1
  • None of these
How many different 5 letter sequences can be made using the letters A,B,C,D with repetition such that the sequence does not include the word BAD?
  • 1024
  • 976
  • 48
  • 678
If 50Cr=Cr then the value of C20+12C21+13C22+14C2+....+150101C513
  • 151101C50
  • 150101C50
  • 151101C49
  • 150101C51
nC0 + 22 . nC12 + 23 . nC23 +.......+ 2n+1 . nCnn+1
  • 3n+1 - 1
  • 3n+11n+1
  • 3n+1n+1
  • 3n+11n
The number of ways in which three girls and ten boys can be seated in two vans, each having numbered seats, three in the front and four at the back is
  • 14C12
  • 14P13
  • 14!
  • 103
The sum nr=1r.2nCr is equal to.
  • n.22n1
  • 22n1
  • 2n1+1
  • none of these
The total no. of six digit numbers x1x2x3x4x5x6 having the property x1<x2x3<x4<x5x6, is equal to
  • 11C6
  • 16C2
  • 17C2
  • 18C2
If nCr denotes the number of combinations of n things taken r things at a time, then the expression nCr+1+nCr1+2nCris
  • n+2Cr+1
  • n+1Cr
  • n+1Cr+1
  • n+2Cr
C1C0+2C2C1+...3C3Cn1=
  • n(n1)2
  • n(n+2)2
  • n(n+1)2
  • (n1)(n2)2
If nr=0(r+2r+1)nCr=2816, then 'n' is 

  • 8
  • 4
  • 6
  • 5
The value 10r=0r10Cr.(2)10r is
  • 10
  • 20
  • 30
  • 300
C1C0+2C2C1+3C3C2+.....+nCnCn1=
  • n(n1)2
  • n(n+2)2
  • n(n+1)2
  • (n1)(n2)2
If (1+x)n=C0+C1x+C2x2+...+Cnxn, then the value of C20+C212+C222+.........+C2nn+1 is 
  • (2n1)!{(n+1)!}2
  • (2n1)!(n+1)!
  • (2n+1)!{(n+1)!}2
  • (2n)!(n+1)!
The number of odd numbers lying between 40000 and 70000 that can be made from the digits 0, 1, 2, 4, 5, 7 if digits can be repeated any number of times is 
  • 1125
  • 1296
  • 766
  • 655
If an=nr=01nCr, then an=nr=0rnCr, equals 
  • (n1)an
  • nan
  • n2an
  • None of these
n2Cr+2n2Cr2+n2Cr2 equals:
  • n+1Cr
  • nCr
  • nCr+1
  • n1Cr
If nC3=nC2, then n is equal to
  • 2
  • 3
  • 5
  • none of these
Ifa=mC2,thenaC2isequalto
  • m+1C4
  • m1C4
  • 3m+2C4
  • 3m+1C4
n1r=0nCrnCr+nCr+1 is equal to :
  • n2
  • n+12
  • n(n+1)2
  • n(n1)2n(n+1)
If 2nC3 : nC2 : : 44:1, then the value of n is 
  • 17
  • 6
  • 11
  • none of these
Total number of 6-digit numbers in which all the odd digits and only odd digits appears, is
  • 52(6!)
  • 6!
  • 12(6!)
  • 32(6!)
nr=1r.nCrnCr1 is equal to:

  • n(n+1)2
  • n(n1)2
  • n(n+1)
  • n(n1)
The value of the expression n+1C2+2[2C2+3C2+4C2+.....+nC2] is
  • n
  • n2
  • n3
  • (n+1)2
The value of 15C8+15C915C615C7 is :
  • -1
  • 0
  • 1
  • None of these
The value of 11C01+11C12+11C23+.......+11C1112 will be 
  • 112(2121)
  • 111(2121)
  • 112(212+1)
  • none of these
How many combinations can be formed of 8 counters marked 1 2 ..... 8 taking 4 at a time there being at lest one odd and even numbered in each combination? 
  • 68
  • 66
  • 64
  • 62
The value of 50C4+6r=156rc3 is 
  • 55c4
  • 55c3
  • 56c3
  • 56c4
Which of the following is not true?
  • Cr=nCnr
  • nCr+nCr1=n+1Cr
  • r.nCr=n.n1Cr1
  • Cr+nCr1=nCr+1
nr=1{r1r1=0nCnrCr12r1}=
  • 4n3n+1
  • 4n3n1
  • 4n3n+2
  • 4n3n
47Cr+5j=1(521)C3=

  • 52C4
  • 52C3
  • 53C4
  • 53C3
If n1Cr=(k23)nCr+1, then k belongs to :
  • (,2]
  • [2,)
  • [3,3]
  • (3,2]
If nr=mrCm=n+1Cm+1, then nr=m(nr+1)rCm is equal to 
  • n+3Cm+2
  • n+2Cm+2
  • n+2Cm+1
  • none of there
 (2nC0)2(2nC1)2+(2nC2)2+(1)2n(2nC2n)2  equals to
  • (2nCn)2
  • (1)2n(2nCn)
  • (1)n(2nCn)
  • (nCn)+(2nCn)
If Cr=nCrand(C0+C1)(C1+C2)......
(Cn1+Cn)=k(n+1)nn! then the value of k is 
  • C0.C1.C2....Cn
  • (C1)2.(C2)2....(Cn)2
  • C1+C2+....+Cn
  • None of these
Total number of four-digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are
  • 216
  • 375
  • 400
  • 720
In a conference hall 10  executive are to be seated on 10 chairs placed left to right. Executive E1 wants to sit to the right of E2 Also, E4 wants to sit the left of E1 and E5 wants to sits to the left of ways E4 Number of ways the executives can be seated on the 10 chairs, is 
  • 10!4!
  • 10!3
  • 10!
  • 3×10!4!
Number of positive integers  n,  less than  17,  for which  n!+(n+1)!+(n+2)!  is an integral multiple of  49  is
  • 2
  • 5
  • 7
  • 9
If 35Cn+7=35C4n2, then the value of n is
  • 3
  • 4
  • 5
  • 6
0:0:1


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