Quiz 5

Let there be m boys and n girls.
Then nC2 = 45 » n(n – 1)=90» n =10
mC2 = 190  » m(m – 1) = 380 » m=20
Number of games played between one boy and one girl
= 10C1 x 20C1= 10 x 20=200
(a)

Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).
Similarly, there are 3 ways to colour each of the remaining stripes.
» The number of ways the flag can be coloured is
4 x (3)⁵ = (12)(3)⁴
(a)

The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.
Total number of ways confusion can arise = 4 x 3=12
Thus, required answer= 81-12=69
(d)

Number of ways for selecting single digit=2
Number of ways for selecting two digit= 2 x 3=6
Number of ways for selecting three digit= 2 x 3 x 3=18
Number of ways for selecting four digit= 2 x 3 x 3 x 3=54
Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162
Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486
Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728
(d)

There are 32 black and 32 white square on a chess-board then number of ways in choosing one  white and one black square on the chess.
32C1 x 32C1= 32 x 32=1024
Number of ways in which square lies in the same row
White square=4
Black  square=4
Number of rows=8
4C1 x 4C1 x 8 = 128
»Number of ways in which square lies in the same column= 128
Total number in which square lie on the same row or same column= 128 + 128=256.
(d)

Ist place of the four letter password can be filled in 11 ways.
IInd place of four letter password can be filled in 10 ways.
IIIrd place of four letter password can be filled in 9 ways.
IVth place of four letter password can be filled in 8 ways.
Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways
(a)

Three letter password from 26 letters  can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.
Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.
(c)

At lease one candidate out of (2n + 1) candidates can be selected in (2ⁿ⁺¹ – 1) ways.
» 2²ⁿ⁺¹ – 1 = 63  » 2²ⁿ⁺¹ = 64 = (2)⁶» n=2.5
Since n cannot be a fraction. Hence n=3.
(a)

Required number of triangles formed
10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045
(c)

The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.
Hence required number of ways = 4!
However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.
Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers  ending in 2 and no number ending with 1. Hence total number of numbers
= 4! + (3) x 3! + (2)3! + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60
(b)