30. Two players P 1 and P 2 play a series of ‘2n’ games.
    Each game can result in either a win or loss for P 1. Total
    number of ways in which P 1 can win the series of these
    games, is equal to

(a)^1
2

() 222 nn− Cn (b)^1

2

() 2222 nn−⋅Cn

(c)^1
2

() 2 nn−^2 Cn (d)^1

2

() 22 nn−⋅^2 Cn

31. Total number of 3 letter words that can be formed
    from the letters of the word ‘SAHARANPUR’, is equal
    to
    (a) 210 (b) 237 (c) 247 (d) 227


32. 15 identical balls have to be put in 5 different
    boxes. Each box can contain any number of balls. Total
    number of ways of putting the balls into box so that
    each box contains atleast 2 balls, is equal to
    (a)^9 C 5 (b)^10 C 5 (c)^6 C 5 (d)^10 C 6


33. Total number of positive integral solutions of the
    equation x 1 · x 2 · x 3 = 60 is equal to
    (a) 27 (b) 54
    (c) 64 (d) none of these


34. Total number of four digit numbers having all
    different digits, is equal to
    (a) 4536 (b) 504 (c) 5040 (d) 720


35. Total number of 5 digit numbers having all different
    digits and divisible by 4 that can be formed using the
    digits {1, 3, 2, 6, 8, 9}, is equal to
    (a) 192 (b) 32 (c) 1152 (d) 384

SOLUTIONS

1. (b) : Let n 1 = x 1 x 2 x 3 x 4 and n 2 = y 1 y 2 y 3 y 4
   n 2 can be subtracted from n 1 without borrowing at any
   stage if xi ≥ yi
   For i = 2, 3, 4; let xi = r(r = 0, 1, 2, ...,9)
   ⇒ yi ≤ r ⇒ yi = 0, 1, 2, ..., r
   That mean yi can be selected in (r + 1) ways.
   Thus, total ways of selecting xi and yi suitably

=+

=

∑()r

r

1

0

9

= 1 + 2 + 3 + ..... + 10 =11 10⋅ =

2

55

For i = 1, let xi ≤ r(r = 1, 2, ...., 9)
⇒ yi ≤ r ⇒ yi = 1, 2, ...., r
That means yi can be selected in ‘r’ ways.
Thus, total ways of selecting x 1 , y 1 suitably

==+++=

⋅

=

=

∑r

r

12 9

910

2

45

1

9

....

Thus, total ways = 45(55)^3

2. (d) : 15 < x 1 + x 2 + x 3 ≤ 20
   ⇒ x 1 + x 2 + x 3 = 16 + r, r = 0, 1, 2, 3, 4.
   Now number of positive integral solutions of

x 1 + x 2 + x 3 = 16 + r is 13 + r + 3 – 1C13 + (^) r,
i.e. 15 + rC 13 + r = 15 + rC 2
Thus required number of solutions
= +

∑
15
2
0
(^4) r
r
C =^15 C 2 +^16 C 2 +^17 C 2 +^18 C 2 +^19 C 2 = 685

3. (b) : If n is odd, then
   3 n = 4λ 1 – 1, 5n = 4λ 2 + 1
   ⇒ 2 n + 3n + 5n is divisible by 4 if n > 2.
   Thus n = 3, 5, 7, 9, ........, 99 i.e., n can take 49 different
   values.
   If n is even, then
   3 n = 4λ 1 + 1, 5n = 4λ 2 + 1
   ⇒ 2 n + 3n + 5n is not divisible by 4, as
   2 n + 3n + 5n will be in the form of 4λ + 2.
   Thus, total number of ways of selecting ‘n’ = 49.


4. (c) : Any integer having less than 4 digits will be in
   the form of xyz.
   If 3 is used exactly once, then the number of ways
   =^3 C 1 · 9^2
   If 3 is used exactly 2 times, then the number of ways
   = (^3 C 2 · 9) · 2
   If 3 is used exactly 3 times, then there is only one such
   number.
   Thus, required number of ways
   = 1 + 2 ·^3 C 2 · 9 +^3 C 1. 9^2 = 298


5. (c) : Total variables if only the alphabet is used is
   equal to 26.
   Total variables if alphabets and digits both are used =
   26 .10.
   ∴ Total variables = 26(1 + 10) = 286


6. (b) : Given numbers can be rearranged as
   1 4 7 .... 3n – 2 → 3 λ – 2 type
   2 5 8 .... 3n – 1 → 3 λ – 1 type
   3 6 9 .... 3n → 3 λ type
   That means we must take two numbers from last or one
   number each from first and second row.
   Total number of ways = nC 2 + nC 1 · nC 1

=nn()−^1 +=n nn−

2

3

2

2 2

7. (a) : Let xW, xR, xB be the number of white balls, red
   balls and blue balls respectively being selected.