Answer : Option 3
Lets do some basic combination math
Here no. of digits = 8
Case – I where ‘0’ occurs at unit place
Unit place can be filled in 1 way (0) ; Ten’s place can be filled in 7 ways
Hundred’s place can be filled in 6 ways ; Thousand’s place can be filled in 5 ways ;
Using fundamental principle of multiplication the required no. = 1×7×6×5= 210;
Case – II When 0 does not occur at unit place
Unit place can be filled in 3 ways (2, 4, 6); Thousand place can be filled in 6 ways; (one of the six digits other than zero); Hundred place can be filled in 6 ways; Ten’s place can be filled in 5 ways. Required number of ways = 3 × 6 × 6 × 5 = 540; Total number of numbers =210 + 540 = 750
How many numbers of four digits greater than 2,400 can be formed with digits 0, 1, 2, 3, 4, 5 & 6; no digit being repeated in any number?
1. 140
2. 480
3. 540
4. 1120
Answer : Option 3
Case – I When 2 occurs at thousand’s place
Thousand’s place can be filled up by 2 in 1 way. Hundred’s place can be filled up by any of the four digits i.e. 4, 5 and 6 in 3 ways; Ten’s place can be filled in 5 ways ; unit’s place can be filled in 4 ways; using fundamental principle of multiplication, the required number = 1 × 3 × 5 × 4 = 60.
Case – II When thousand’s place can be occupied by any of the digits out of 3, 4, 5 and 6; thousand’s place can be filled in 4 ways; Hundred’s place can be filled in 6 ways; Ten’s place can be filled in 5 ways. Unit’s place can be filled in 4 ways. Hence, the total number of ways are: 4 x 6 x 5 x 4 = 480. So, the required numbers = 60 + 480 = 540.
There are 20 books of which 4 are single volume and the other are books of 8, 2 and 6 volumes respectively. In how many ways can all these books be arranged on a shelf so that volumes of the same book are not separated?
1. 2!3!4!5!
2. 3!4!5!6!
3. 4!5!6!7!
4. 8! 7! 6! 2!
Answer : Option 4
Here volumes of the same book are not to be separated i.e. all the volumes of the same book are to be kept together. Regarding all volumes of the same book as one book, we have only 4 + 1 + 1 + 1 = 7 books. These 7 books can be arranged in 7! ways. Volumes of book having 6 volumes can be arranged in 6! ways. Volumes of book having 8 volumes can be arranged in 8! ways.
Volumes of book having 2 volumes can be arranged in 2! ways. ∴ Required number = 7! 8! 6! 2!.
How many different words can be formed with the letters of the word ‘FAMILY’ when vowels occupy even places?
1. 18
2. 36
3. 72
4. 144
Answer : Option 4
There are 6 letters in the word FAMILY and no letter is repeated. There are two vowels ‘A’ and ‘I’; Places are 1st, 2nd, 3rd, 4th, 5th, 6th; Now there are 3 even places for 2 vowels. ∴ 2 vowels can be arranged in 3P2 = 3! = 6 ways. Four consonants can be arranged in the remaining places in = 4! =24 ways. Required Number = 6 × 24 = 144.
A round table conference is to be held among 25 delegates from 25 countries. In how many ways can they be seated if two particular delegates are always to sit together?
1. 23!
2. 2! ×23!
3. 3! ×23!
4. None of these
Answer : Option 2
Treating 2 particular delegates who are to sit together as one person, we have only 24 persons. These 24 persons can be seated at a round table in 23! Ways. But 2 particular persons can be arranged among themselves in 2! ways. ∴ Required no =2!×23!
Given 5 line segments of lengths 2, 3, 4, 5, 6 and 7 units. Then the number of triangles that can be formed by joining these lines is
1. 6C3
2. 6C3 - 7
3. 6C3 - 5
4. 6C3 - 1
Answer : Option 2
We know that in any triangle the sum of two sides is always greater than the third side. ∴ the triangle will not be formed if we select segments of lengths (2, 3, 5) , (2, 3, 6), (2,3,7), (3,4,7),(2,4,7), (2,5,7) and (2, 4, 6). Hence number of triangles formed = 6C3 - 7
Find the number of divisors of 43200.
1. 48
2. 60
3. 72
4. 84
Answer : Option 4
If a number N = ax . by . cz then no of divisors = (x+1)(y+1)(z+1)
Here 43,200 = 26 × 33 × 52 ∴ Number of Divisors = (6 + 1)(3 + 1)(2 + 1) = 84
A gentleman has 5 friends to invite. In how many ways can he send invitation cards to them if he has four servants to carry the cards?
Answer : Option 3
Here each card can be carried by any of the four servants.∴ Required number = 4 × 4 × 4 × 4 × 4 = 45 = 1024
In how many ways can 5 boys and 4 girls be seated in a row, so that they alternate?
1. 5!
2. 5!×2!
3. 4!×5!
4. None of these
Answer : Option 3
B G B G B G B G B. There are five places i.e. 1st , 3rd, 5th ,7th and 9th for five boys. Five boys can be seated in 5! ways. Again there are four places i.e. 2nd, 4th, 6th, 8th for four Girls. ∴ four girls can be seated in 4! ways ;
∴ Required number = 5!4!
In how many ways can 5 boys and 5 girls can be seated in a row so that boys and girls are placed alternately?
1. 5!
2. 5!×2!
3. 2×5! × 5!
4. None of these
Answer : Option 3
The 5 boys and 5 girls can be seated in 5!5! ways. There are further two cases when the arrangement starts with boy or girl. Thus the required number of arrangements is 2 × 5!×5!
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