3
Q:

# The number of signals that can be generated by using 5 differently coloured flags, when any number of them may be hoisted at a time is:

 A) 235 B) 253 C) 325 D) None of these

Explanation:

Required number of signals = $\inline&space;^{5}\textrm{P}_{1}+^{5}\textrm{P}_{2}+^{5}\textrm{P}_{3}+^{5}\textrm{P}_{4}+^{5}\textrm{P}_{5}$

= 5 + 20 + 60 + 120 + 120 = 325

Q:

The number of permutations of the letters of the word 'MESMERISE' is  ?

 A) 9!/(2!)^{2}x3! B) 9! x 2! x 3! C) 0 D) None

Explanation:

n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.
Number of arrangements = $\inline&space;\fn_jvn&space;\small&space;\frac{9!}{(2!)^{2}x3!}$

0 50
Q:

A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ?

 A) ²²C₁₀ + 1 B) ²²C₉ + ¹⁰C₁ C) ²²C₁₀ D) ²²C₁₀ - 1

Explanation:

The total number of ways of forming the group of ten representatives is ²²C₁₀.
The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way
The required number of ways = ²²C₁₀ - 1

1 44
Q:

A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ?

A)  $\inline \fn_jvn \small 2^{4} - 1$     B) $\inline \fn_jvn \small 2^{4}(2^{5}-1)$     C) $\inline \fn_jvn \small (2^{4}-1)(2^{3}-1)2^{5}$   D) None

 A) A B) B C) C D) D

Explanation:

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.

Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.

Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.

Hence, number of ways in which we can select the black balls

= 4C1 + 4C2 + 4C3 + 4C4
= $\inline \fn_jvn \small 2^{4} - 1$  ........(A)

Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.

Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls

Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=$\inline \fn_jvn \small 2^{3}-1$ ........(B)

Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.

Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + … + 5C5
= $\inline \fn_jvn \small 2^{5}$..............(C)

From (A), (B) and (C), required number of ways
=  $\inline \fn_jvn \small (2^{4}-1)(2^{3}-1)2^{5}$

1 47
Q:

How many three digit numbers 'abc' are formed where two of the three digits are same ?

 A) 252 B) 648 C) 243 D) 900

Explanation:

Digits are 0,1,2,3,4,5,6,7,8,9
so no. of digits are 10
first all possible case => 9(0 excluded) x 10 x 10 = 900
second no repetition allowed =>9 x 9 x 8 = 648
third all digits are same => 9 (111,222,333,444,555,666,777,888,999)
Three digit numbers where two of the three digits are same = 900 - 648 - 9 = 243 ;

1 94
Q:

Find the sum of the all the numbers formed by the digits 2,4,6 and 8 without repetition. Number may be of any of the form like 2,24,684,4862 ?

 A) 133345 B) 147320 C) 13320 D) 145874

Explanation:

Sum of 4 digit numbers = (2+4+6+8) x 3P3 x (1111) = 20 x 6 x 1111 = 133320
Sum of 3 digit numbers = (2+4+6+8) x 3P2 x (111) = 20 x 6 x 111 = 13320
Sum of 2 digit numbers = (2+4+6+8) x 3P1 x (11) = 20 x 3 x 11 = 660
Sum of 1 digit numbers = (2+4+6+8) x 3P0 x (1) = 20 x 1 x 1 = 20