Topic Question Set


Q 41 :

The number of 4-letter words (with or without meaning) that can be formed from the eleven letters of the word 'EXAMINATION' is ________.



(2454)

The word EXAMINATION has 2N, 2A, 2I, E, X, M, T, O

Case I:  If all letters are different, then number of words

=P48=8!4!=8×7×6×5=1680

Case II:  If 2 letters are same and 2 are different, then number of words

=C13·C27·4!2!=3×21×12=756

Case III:  If 2 letters are same of one kind and 2 letters are same of different kind, then number of words

=C23·4!2!2!=3×6=18

   Total number of words =1680+756+18=2454



Q 42 :

The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word 'SYLLABUS' such that two letters are distinct and two letters are alike, is ________.



(240)

The word SYLLABUS consists of

2S, 2L, 1Y, 1A, 1B, 1U

Required number of ways=C12×C25×4!2!

                                                     =2×10×12=240



Q 43 :

The value of C916+C1016-C616-C716 is

  • 1

     

  • C1017

     

  • C317

     

  • 0

     

(4)

We have,  C916+C1016-C616-C716

=C1017-C717                  ( Crn+Cr-1n=Crn+1)

=C1017-C1017                       ( Crn=Cn-rn)

=0



Q 44 :

There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicket keepers. The number of ways a team of 11 players can be selected from them so as to include at least 4 bowlers, 5 batsmen and 1 wicket keeper is ________.



(777)

Total number of players = 15

Number of bowlers = 6

Number of batsmen = 7

Number of wicket keepers = 2

   Total number of ways for which at least 4 bowlers, 5 batsmen and 1 wicket keeper is to be selected

=C46C57C12+C46C67C12+C56C57C12

=315+210+252=777



Q 45 :

The value of  k=06C351-k is equal to

  • C351-C345

     

  • C452-C445

     

  • C352-C345

     

  • C451-C445

     

(2)

k=06C351-k=C351+C350+C349+C348+C347+C346+C345

As we know,  Crn+Cr-1n=Crn+1    Cr-1n=Crn+1-Crn

So,  C351=C452-C451,    C350=C451-C450   and   C345=C446-C445

  k=06C351-k=C452-C451+C451-C450++C447-C446+C446-C445

       =C452-C445



Q 46 :

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is ________.



(3483638676)

Required number of ways

=C03320-C13220+C23120

=320-3×220+3=3(319-220+1)=3483638676



Q 47 :

The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is ________.



(432)

Vowels are U, I, E, E and consonants are N, V, R, S.

So, there are 4 vowels and 4 consonants.

When both vowels are different,

Number of ways =C23×C24×4!=432



Q 48 :

A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is ________.



(546)

5 language courses and 7 non-language courses.

He can select:

(i)    0 language courses and 5 non-language courses in C05×C57=21 ways

(ii)   1 language course and 4 non-language courses in C15×C47=175 ways

(iii)   2 language courses and 3 non-language courses in  C25×C37=350 ways

Required number of ways =21+175+350 =546 ways.



Q 49 :

The rank of the word 'SUCCESS' in the dictionary is ________.



(331)

The alphabetical order is C, E, S, U. The number of words beginning with C is 6!3!=120 ways, and those beginning with E is 6!2!3!=60 ways.

Then come words beginning with SC, numbering 5!2!=60, SE, numbering 5!2!2!=30, and SS, numbering 5!2!=60.

After which, comes the word SUCCESS.

Thus, the rank of SUCCESS is 120+60+60+30+60+1=331



Q 50 :

The rank of the word 'ARRANGE' in the dictionary is ________.



(342)

The alphabetical order is A, E, G, N, R. The number of words beginning with AA is 5!2!=60, beginning with AE is 5!2!=60, beginning with AG is 5!2!=60, beginning with AN is 5!2!=60. Then come words beginning with ARA is 4!=24, beginning with ARE is 4!=24, beginning with ARG is 4!=24, and beginning with ARN is 4!=24, then come words beginning ARRAE is 2!=2, beginning with ARRAG is 2!=2, then next words are ARRANEG and ARRANGE.

Thus, the rank of ARRANGE is 60+60+60+60+24+24+24+24+2+2+2=342



Q 51 :

The number of four-letter words that can be formed using the letters of the word BARRACK is ________.



(270)

Given word : BARRACK

Case 1:  If all four letters are different, then number of words =C45×4!=120

Case 2:  If two letters are of same kind and other are different, then number of words =C12×C24×4!2!=144

Case 3:  If first two are same and second two are also same letters, then number of words =4!2!2!=6

  Total number of words =120+144+6 =270



Q 52 :

Relatives of a man comprise 4 ladies and 3 gentlemen, and his wife has also 7 relatives: 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of the man's relatives and 3 of the wife's relatives?



(485)

3 gentlemen and 3 ladies may be chosen as follows:

Case Man Wife
Case I 3(L) 3(G)
Case II 2(L), 1(G) 2(G), 1(L)
Case III 1(L), 2(G) 2(L), 1(G)
Case IV 3(G) 3(L)

Required number of ways of inviting 3 ladies and 3 gentlemen to the party

=C34×C34+(C24×C13)(C13×C24)+(C14×C23)(C23×C14)+C33×C33

=16+324+144+1=485