Topic Question Set


Q 21 :    

Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is :           [2025]

  • 13

     

  • 23

     

  • 14

     

  • 12

     

(4)

We have, A E G R D N

Total = 6!

Number of cases when A and E are in order = C26×4!

Probability, when A and E are in order = C26×4!6!=12

Probability when A and E are in order = 112=12



Q 22 :    

For n2, let Sn denote the set of all subsets of {1 2, ..., n} with no two consecutive numbers. For example {1,3,5}S6 but {1,2,4}S6. Then n(S5) is equal to __________.          [2025]



13

Let A = {1, 2, 3, 4, 5, ..., n}

Number of subsets having r elements such that no two are consecutive = Crnr+1

For n = 5, number of ways = Cr6r

Subsets having no elements = 1 i.e.ϕ

Subsets having exactly 1 element = C15 = 5 i.e., {1}, {2}, {3}, {4}, {5}

Subsets having exactly 2 element = C24 = 6 i.e., {1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {3, 5}

Subsets having exactly 3 element = C33 = 1 i.e., {1, 3, 5}

  n(S5) = 1 + 5 + 6 + 1 = 13



Q 23 :    

The number of singular matrices of order 2, whose elements are from the set {2, 3, 6, 9}, is __________.          [2025]



36

For any singular matrix [acbd]; we ave

|acbd|=adbc=0  ad=bc

Case I : Exactly one number is used

                  All like, then required number = C14 = 4

Case II : Exactly two number i.e., (a, a, a, b or a, a, b, b) are used, then rrequired number

                 C24×2×2=24

Case III : Exactly three numbers i.e., (a, ab, c) are used, then none will be singular.

Case IV : Exatly four number i.e., (a, b, c, d) are used, then ad = bc i.e.,

                2×9=3×6  C14×2!=8 

Total number of matrices = 36.

 



Q 24 :    

If r=05C2r+1112r+2=mn, gcd (m, n) = 1, then mn is equal to __________.          [2025]



2035

We know that,

C0+C12+C23+......+Cnn+1+2n+11(n+1)          ... (i)

C0C12+C23......+(1)nCnn+1+1n+1          ... (ii)

Subtracting (ii) from (i), we get

2[C12+C34+......]=2n+11(n+1)1(n+1)

Put n = 11, we get

2[C12+C34+......+C1112]=212112112

 C12+C34+......+C1112=12[212212]          ... (iii)

Now, r=05C2r+1112r+2=mn

 C1112+C3114+C5116+......+C111112=mn

 12[212212]=mn          [From (iii)]

 211112=mn  mn=204712

   mn = 2047 – 12 = 2035.



Q 25 :    

If r=130r2(Cr30)2Cr130=α×229, then α is equal to __________.          [2025]



465

r=130r2(Cr30)2Cr130=r=130r2(31rr)·30!r!(30r)!

=r=130(31r)30!(r1)!(30r)!=30r=130(30r+1)×C30r29

=30(r=130(30r)×C30r29+r=130C30r29)

=30(29×228+229)

      [  C1n+2C2n+3C3n+4C4n+...+nCnn=n2n1 and C1n+C2n+C3n+...Cnn=2n1]

=30(29+2)228=15×31×229=465(229)

  α=465.



Q 26 :    

Let S={p1,p2,......,p10} be the set of first ten prime numbers, Let A=SP, where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs (x,y), xS, yA, such that x divides y, is __________.          [2025]



5120

S = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

P={2×3,2×3×5,...}

A=SP={2,2×3,...,3;3×5,...}

For x = 2, the value of y can be

1+C19+C29+C39+...+C99=29

Similarly, for x = 3, 5, 7, 11, ...; y can be 29

   Required number of ordered pair = 10×(29)

                                                             = 10×512=5120



Q 27 :    

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is __________.          [2025]



1405

(i) Single letter is used, then number of words = 5

(ii) Two distinct letters are used, then number of words

              =C25×(6!2!4!×2+6!3!3!)=10(30+20)=500

(iii) Three distinct letters are used, then number of words

             =C35×6!2!2!2!=900

   Total number of words = 1405.