Mathmetics
ISBT Mathmetics for all banking PO,Clerk,IBPS PO,Railway,SSC,IAS,OAS Exams
Q1. | How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed ? |
| 1) | 30 | 2) | 48 |
| 3) | 60 | 4) | 120 |
| 5) | None of these | ||
Answer : 60
Explanation :
Explanation :
Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is not allowed,
The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.
Therefore, The total number of possible 3-digit numbers = 5 × 4 × 3=60
View AnswerQ2. | How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed ? |
| 1) | 60 | 2) | 90 |
| 3) | 120 | 4) | 240 |
| 5) | None of these | ||
Answer : 120
Explanation : Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is allowed,
The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.
Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125
Explanation : Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.
Now when repetition is allowed,
The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.
Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125
View AnswerQ3. |
|
| 1) | 26 | 2) | 27 |
| 3) | 41 | 4) | 56 |
| 5) | None of these | ||
Answer : 41
View AnswerQ4. |
|
| 1) | 2420 | 2) | 2880 |
| 3) | 3260 | 4) | 4840 |
| 5) | None of these | ||
Answer : 2880
Explanation :
Explanation :
Treating the group of four vowels (E, I O and U) as one distinct letter then total number of letters will be 8 - 4 + 1 = 5 [i.e. P, R, V, S and (E, I, O, U)].
The five letters can be arranged in 5! = 120 ways The vowels among themselves can be arranged in 4! = 24 ways,
The letters of the word ‘PREVIOUS’ can be arranged in 120 × 24 = 2880 ways such that vowels always come together.
View AnswerQ5. | There are 5 men and 3 women to form a committee of 5-members where at least 3 men are there. How many way they can be selected ? |
| 1) | 42 | 2) | 45 |
| 3) | 46 | 4) | 54 |
| 5) | None of these | ||
Answer : 46
Explanation : Total ,5 men and 3 women i.e. 8 persons.
Explanation : Total ,5 men and 3 women i.e. 8 persons.
So, 3 men can selected from 5 men.
They can selected in (5C3 x 3 C2 ) or (5C4 x 3 C1 ) or (5C5 x 3 C0 )
Total numbers of way to select = [(5 x 4 )/2 x 3 ] +[(5 ) x 3 ] + [(1x 1) ] = 46
View AnswerQ6. | There are 5 professors from IIT Delhi and 3 professors from IIT Bombay who are sitting together to discuss a matter at a round table. In how many ways can they sit around the table so that no two professors from IIT Bombay are together ? |
| 1) | 600 | 2) | 840 |
| 3) | 1200 | 4) | 1440 |
| 5) | None of these | ||
Answer : 1440
Explanation : Given, no professors of IIT-B together. Suppose, we first arrange 5 prof. IIT-D and after that we can arrange the IIT-B in space between IIT-D.
Explanation : Given, no professors of IIT-B together. Suppose, we first arrange 5 prof. IIT-D and after that we can arrange the IIT-B in space between IIT-D.
So, numbers of way of arranging the IIT-Bin the circle = (5-1)! = 24 ways.
Numbers of way of arranging IIT-B will be by placing them in 5 space that are formed between the IIT-D.
This can be arranged in 5p3 = 60 ways.
therefore, total number of ways = 24 x 60 =1440
View AnswerQ7. |
|
| 1) | 32 | 2) | 48 |
| 3) | 56 | 4) | 64 |
| 5) | None of these | ||
Answer : 64
Explanation : Out of total 10 balls,
Explanation : Out of total 10 balls,
View AnswerQ8. |
|
| 1) | 266 | 2) | 5040 |
| 3) | 11760 | 4) | 86400 |
| 5) | None of these | ||
Answer : 11760
Explanation : From 8 men and 10 women , the committee can be formed in following ways.
Explanation : From 8 men and 10 women , the committee can be formed in following ways.
View AnswerQ9. |
|
| 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 AnswerQ10. |
|
| 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 Answer