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21. The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?
Now, with A in the beginning, the remaining letters can be permuted in 5! ways.
Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.
With L in the beginning, the first word will be LABORU, the second will be LABOUR.
Hence, the rank of the word LABOUR is,
5! + 5! + 2 = 120 + 120 + 2 = 242
Note: 5! = 5 × 4 × 3 × 2 × 1 = 120
Solution:
The order of each letter in the dictionary is ABLORU.Now, with A in the beginning, the remaining letters can be permuted in 5! ways.
Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.
With L in the beginning, the first word will be LABORU, the second will be LABOUR.
Hence, the rank of the word LABOUR is,
5! + 5! + 2 = 120 + 120 + 2 = 242
Note: 5! = 5 × 4 × 3 × 2 × 1 = 120
22. A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?
Girls can be arranged in 6! ways.
Total number of ways in which all the students can be arranged,
= 2! × 18! × 6!
= 18! × 1440
Note:
N! = N × (N - 1) × (N - 2) × . . . . × 1
So,
18! = 18 × 17 × 16 × 15 . . . . . . . . × 1
Solution:
Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways.Girls can be arranged in 6! ways.
Total number of ways in which all the students can be arranged,
= 2! × 18! × 6!
= 18! × 1440
Note:
N! = N × (N - 1) × (N - 2) × . . . . × 1
So,
18! = 18 × 17 × 16 × 15 . . . . . . . . × 1
23. If find n?
(n+1)P3 = (n+1) × n × (n–1)
Now,
5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1)
Or, 5(n−2) = 4(n+1)
Or, 5n − 10 = 4n + 4
Or, 5n − 4n = 4 + 10
Hence, n = 14
Note:nPr =
Solution:
nP3 = n × (n–1) × (n–2)(n+1)P3 = (n+1) × n × (n–1)
Now,
5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1)
Or, 5(n−2) = 4(n+1)
Or, 5n − 10 = 4n + 4
Or, 5n − 4n = 4 + 10
Hence, n = 14
Note:nPr =
24. Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:
= Total number of words - total number of words in which no letter is repeated.
= 105 – 16P5
= 100000 − 30240
= 69760
Solution:
Number of words which have at least one letter replaced,= Total number of words - total number of words in which no letter is repeated.
= 105 – 16P5
= 100000 − 30240
= 69760
25. 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
The various combinations of chairs that ensure that no two men are sitting together are listed.
(1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.
(1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways.
(1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 8, 10, 12) is also one of the combinations.
Hence, 16 such combinations exist.
In case of each these combinations we can make the four men inter arrange in 4! ways.
Hence, the required result =16 × 4! = 384
Solution:
Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied.1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
The various combinations of chairs that ensure that no two men are sitting together are listed.
(1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.
(1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways.
(1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 8, 10, 12) is also one of the combinations.
Hence, 16 such combinations exist.
In case of each these combinations we can make the four men inter arrange in 4! ways.
Hence, the required result =16 × 4! = 384
26. How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?
A _ _ _ _ T.
The remaining 4 positions can be filled in 4! Ways by the remaining words (S, C, E, N, T).
Hence by rearranging the letters of the word ASCENT we can form,
1 x 4! = 4! Words.
Solution:
As A and T should occupy the first and last position, the first and last position can be filled in only one following way.A _ _ _ _ T.
The remaining 4 positions can be filled in 4! Ways by the remaining words (S, C, E, N, T).
Hence by rearranging the letters of the word ASCENT we can form,
1 x 4! = 4! Words.
27. If 6Pr = 360 and If 6Cr = 15, find r ?
6Pr = 15 × r!
360 = 15 × r!
r! = = 24
r! = 4 × 3 × 2 × 1
⇒ r! = 4!
Therefore, r = 4
Solution:
nPr = nCr × r!6Pr = 15 × r!
360 = 15 × r!
r! = = 24
r! = 4 × 3 × 2 × 1
⇒ r! = 4!
Therefore, r = 4
28. In how many ways can six different rings be worn on four fingers of one hand?
So, each ring may be worn in 4 different ways.
∴ 6 rings may be worm in (4×4×4×4×4×4) = 46 = 4096 ways.
Solution:
Each ring may be worn in any of 4 fingers.So, each ring may be worn in 4 different ways.
∴ 6 rings may be worm in (4×4×4×4×4×4) = 46 = 4096 ways.
29. There are 7 non-collinear points. How many triangles can be drawn by joining these points?
There are 7 non-collinear points.
The number of triangles formed,
Solution:
A triangle is formed by joining any three non-collinear points in pairs.There are 7 non-collinear points.
The number of triangles formed,
30. From 6 men and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done, if the committee is to include at least one lady?
(1 lady + 4 gents) or (2 ladies + 3 gents) or (3 ladies + 2 gents) or (4 ladies + 1 gents) or (5 ladies + 0 gents).
Total number of possible arrangements,
= (4C1 × 6C4) + (4C2 × 6C3) + (4C3 × 6C2) + (4C4 × 6C1)
= 60 + 120 + 60 + 6
= 246
Solution:
To committee can be formed in the following ways,(1 lady + 4 gents) or (2 ladies + 3 gents) or (3 ladies + 2 gents) or (4 ladies + 1 gents) or (5 ladies + 0 gents).
Total number of possible arrangements,
= (4C1 × 6C4) + (4C2 × 6C3) + (4C3 × 6C2) + (4C4 × 6C1)
= 60 + 120 + 60 + 6
= 246