23. (a) : Let the number be n = x 1 x 2 x 3
    Since x 1 + x 2 + x 3 is even. That means there are following
    cases :
    (i) x 1 , x 2 , x 3 all are even → 4 · 5 · 5 = 100 ways
    (ii) x 1 is even and x 2 , x 3 are odd → 4 · 5 · 5 = 100 ways
    (iii) x 1 , x 2 are odd and x 3 is even → 5 · 5· 5 = 125 ways
    (iv) x 1 is odd, x 2 is even and x 3 is odd → 5 · 5 · 5
    = 125 ways
    Total ways = 100 + 100 + 125 + 125 = 450.


24. (b) : In this case, one child gets no toy and one
    gets 2 toys and all remaining children get one each.
    Corresponding number of ways

=

−

n ⋅ = ⋅

n

! nn Cn

!( )!

!!

22 2

25. (d) : Total permutations = 0 + k + k^2 + k^3 + ..... + kr

= −

−

kk

k

()r

()

1

1

26. (b) : Let the formed number is x 1 x 2 x 3 x 4
    Clearly, x 1 ≥ 3.
    Thus total number of such numbers
    = 4 · 5 · 4 · 3 = 240


27. (c) : Let the number of students be n, then total
    number of times the teacher goes to zoo is equal to nC 3
    and total number of times a particular student goes to
    the zoo is equal to n – 1 C 2
    Thus nC 3 – n – 1C 2 = 84

⇒ nn()( )−−n −n−−n =

!

12 ()( )

3

12

2

84

⇒ n(n – 1) (n – 2) – 3(n – 1) (n – 2) = 504
⇒ (n – 1) (n – 2) (n – 3) = 504
⇒ (n – 1) (n – 2) (n – 3) = 9 · 8 · 7
⇒ n = 10

28. (a) : Matches whose predictions are correct can be
    selected in^20 C 10 ways. Now each wrong prediction can
    be made in 2 ways.
    Thus total ways =^20 C 10 ⋅ 210.


29. (c) : Let S 1 and S 2 refuse to be together and S 3 and
    S 4 want to be together only.
    Total ways when S 3 and S 4 are selected
    = (^8 C 2 +^2 C 1 ·^8 C 1 ) = 44
    Total ways when S 3 and S 4 are not selected
    = (^8 C 4 +^2 C 1 ·^8 C 3 ) = 182
    Thus total ways = 44 + 182 = 226.
    30. (a) : ‘P 1 ’ must win atleast (n + 1) games.
    Let ‘P 1 ’ wins n + r games (r = 1 , 2 , ...., n)

Corresponding ways =^2 nCn (^) + (^) r
Total ways =^2
1
n nr
r
n
C+

∑
=^2 nCn (^) + 1 +^2 nCn (^) + 2 + .... +^2 nC 2 n

2 −
2
22 nn
Cn= (^1) −
2
() 222 nnCn

31. (c) : 1S, 3A, 1H, 2R, 1N, 1P, 1U when all letters are
    different.
    Corresponding ways =^7 C 3 · 3! =^7 P 3 = 210
    When two letters are of one kind and other is different.
    Corresponding ways =^21 ⋅⋅^613 =
    2

CC! 36

!

When all letters are alike, corresponding ways

= 210 + 36 + 1 = 247

32. (a) : Let the balls put in the box are x 1 , x 2 , x 3 , x 4 and
    x 5.
    We m u s t h a v e
    x 1 + x 2 + x 3 + x 4 + x 5 = 15, xi ≥ 2
    ⇒ (x 1 – 2) + (x 2 – 2) + (x 3 – 2) + (x 4 – 2) + (x 5 – 2)
    = 5
    ⇒ y 1 + y 2 + y 3 + y 4 + y 5 = 5, yi = xi – 2 ≥ 0
    Total number of ways is simply equal to number of non-
    negative integral solutions of the last equation, which is
    equal to 5 + 5 – 1C 5 =^9 C 5.


33. (b) : x 1 · x 2 · x 3 = 2^2 · 3 · 5
    ⇒ Total positive integral solutions = 54


34. (a) : Let the number be x 1 x 2 x 3 x 4.
    Then x 1 can be chosen in 9 ways. x 2 can be chosen in 9
    ways. Similarly x 3 and x 4 can be chosen in 8 and 7 ways
    respectively.
    ∴ Total number of such numbers
    = 9 · 9 · 8 · 7 = 4536


35. (a) : A number is divisible by four, if the last two
    digits are divisible by four. In this case last two digits
    can be 12, 16, 28, 32, 36, 68, 92 or 96.
    Total number of such numbers = 8 · (^4 C 3 · 3!) = 192.
    ””