Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q511. |
|
| 1) | 60 | 2) | 120 |
| 3) | 180 | 4) | 240 |
| 5) | None of these | ||
Answer : 120
Explanation :
Explanation :
The word 'DELHI' has 5 letters and all these letters are different.
Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120
View AnswerQ512. |
|
| 1) | 60 | 2) | 120 |
| 3) | 240 | 4) | 360 |
| 5) | None of these | ||
Answer : 360
Explanation :
Explanation :
The word 'LEADER' has 6 letters.
But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.
Hence,number of ways to arrange these letters = 6!2!=6×5×4×3×2×12×1 = 360
View AnswerQ513. |
|
| 1) | 8750 | 2) | 10420 |
| 3) | 11760 | 4) | 12420 |
| 5) | None of these | ||
Answer : 11760
Explanation :
Explanation :
We need to select 5 men from 8 men and 6 women from 10 women
Number of ways to do this = 8C5 x 10C6
= 8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]
= (8×7×63×2×1) (10×9×8×74×3×2×1)
= 56 x 210 = 11760
View AnswerQ514. |
|
| 1) | 2100 | 2) | 24400 |
| 3) | 21300 | 4) | 25200 |
| 5) | None of these | ||
Answer : 25200
Explanation :
Explanation :
Number of ways of selecting 3 consonants out of 7 = 7C3
Number of ways of selecting 2 vowels out of 4 = 4C2
Number of ways of selecting 3 consonants out of 7 and 2
vowels out of 4 = 7C3 x 4C2
=(7×6×5/3×2×1)×(4×3/2×1)=210
It means that we can have 210 groups where each group
contains total 5 letters(3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120
Hence, Required number of ways = 210 x 120 = 25200
View AnswerQ515. |
|
| 1) | 100800 | 2) | 120960 |
| 3) | 240150 | 4) | 4989600 |
| 5) | None of thesae | ||
Answer : 120960
Explanation :
Explanation :
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8! / (2!)(2!)= 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4! /2! = 12.
So, required number of words = (10080 x 12) = 120960.
View AnswerQ516. |
|
| 1) | 42 | 2) | 48 |
| 3) | 63 | 4) | 90 |
| 5) | None of these | ||
Answer : 63
Explanation : The required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = (7 x 6 )/(2 x 1) x 3 = 63.
Explanation : The required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = (7 x 6 )/(2 x 1) x 3 = 63.
View AnswerQ517. |
|
| 1) | 24 | 2) | 32 |
| 3) | 36 | 4) | 48 |
| 5) | None of these | ||
Answer : 36
Explanation :
Explanation :
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
View AnswerQ518. |
|
| 1) | 6 | 2) | 10 |
| 3) | 12 | 4) | 20 |
| 5) | None of thse | ||
Answer : 20
Explanation :
Explanation :
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
So, required number of numbers = (1 x 5 x 4) = 20.
View AnswerQ519. |
|
| 1) | 810 | 2) | 1440 |
| 3) | 2880 | 4) | 50400 |
| 5) | 57600 | ||
Answer : 50400
Explanation :
Explanation :
In the word 'CORPORATION', we consider the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3!= 20 ways.
So, required number of ways = (2520 x 20) = 50400.
View AnswerQ520. |
|
| 1) | 360 | 2) | 480 |
| 3) | 720 | 4) | 1280 |
| 5) | None of these | ||
Answer : 720
Explanation :
Explanation :
The word 'LEADING'' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
So, required number of ways = (120 x 6) = 720.
View Answer