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- Question
In how many different ways can the letters of the word 'ABYSMAL' be arranged ?
Options- A. 5040
- B. 3650
- C. 4150
- D. 2520
- Correct Answer
- 2520
ExplanationTotal number of letters in the word ABYSMAL are 7
Number of ways these 7 letters can be arranged are 7! ways
But the letter is repeated and this can be arranged in 2! ways
Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.
- 1. In how many different ways can the letters of the word 'POVERTY' be arranged ?
Options- A. 2520
- B. 5040
- C. 1260
- D. None Discuss
- 2. In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?
Options- A. 18000
- B. 17280
- C. 17829
- D. 18270 Discuss
- 3. Find the value of 'n' for which the nth term of two AP'S: 15,12,9.... and -15,-13,-11...... are equal?
Options- A. n = 2
- B. n = 5
- C. n = 29/5
- D. n = 1 Discuss
- 4. There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?
Options- A. 1170
- B. 1200
- C. 720
- D. 360 Discuss
- 5. There are 7 non-collinear points. How many triangles can be drawn by joining these points?
Options- A. 10
- B. 30
- C. 35
- D. 60 Discuss
- 6. Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?
Options- A. 2400
- B. 6400
- C. 3600
- D. 7200 Discuss
- 7. If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?
Options- A. 32
- B. 24
- C. 72
- D. 36 Discuss
- 8. How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ?
Options- A. 521
- B. 720
- C. 420
- D. 225 Discuss
- 9. The number of ways in which 8 distinct toys can be distributed among 5 children?
Options- A. 5P8
- B. 5^8
- C. 8P5
- D. 8^5 Discuss
- 10. In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?
Options- A. 720
- B. 1440
- C. 1800
- D. 3600 Discuss
Permutation and Combination problems
Search Results
Correct Answer: 5040
Explanation:
The 7 letters word 'POVERTY' be arranged in ways = 7! = 5040 ways.
Correct Answer: 17280
Explanation:
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280
Correct Answer: n = 29/5
Explanation:
Given are the two AP'S:
15,12,9.... in which a=15, d=-3.............(1)
-15,-13,-11..... in which a'=-15 ,d'=2.....(2)
now using the nth term's formula,we get
a+(n-1)d = a'+(n-1)d'
substituting the value obtained in eq. 1 and 2,
15+(n-1) x (-3) = -15+(n-1) x 2
=> 15 - 3n + 3 = -15 + 2n - 2
=> 12 - 3n = -17 + 2n
=> 12+17 = 2n+3n
=> 29=5n
=> n= 29/5
Correct Answer: 1170
Explanation:
Possibilities Bowlers Batsmen Number of ways
6 9
1 4 7
2 5 6
3 6 5
= 15 x 36 = 540
= 6 x 84 = 504
= 1 x 126 = 126
Total = 1170
Correct Answer: 35
Explanation:
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points
The number of triangles formed = = 35
Correct Answer: 3600
Explanation:
Let the beds be numbered 1 to 7.
Case 1 : Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5´5!ways = 600 ways.
Case 2 : Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju?s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.
The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways
Correct Answer: 32
Explanation:
The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24+6+2= 32
Correct Answer: 720
Explanation:
Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240
Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.
Correct Answer: 5^8
Explanation:
As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are ways to distribute the toys.
Hence, it is and not .
Correct Answer: 1440
Explanation:
Given word is THERAPY.
Number of letters in the given word = 7
Number of vowels in the given word = 2 = A & E
Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is
6! x 2! = 720 x 2 = 1440.
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