# number of arrangements with elements repetition or with no repetition

Now, if we have to determine the number of 3-letter words, with or without meaning, which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN, NUB, …, etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words =6×5×4=120 (by using multiplication principle).
If the repetition of the letters was allowed, the required number of words would be 6×6×6=216.

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

Q1. How many possible choices can be made in a multiple choice quiz if there are 4 questions each with 3 answers?
Solution:
Four ‘boxes’ of 3 ∴ =3×3×3×3=81 possible choices.

Q2. In how many ways, 5 distinct letters can be posted in 7 letter boxes?
a. 57 b. 7C5 C. 75 d. 7P5
correct: c, Solution:
1st letter can be posted in 7 different ways. 2nd letter in 7 ways … 5th letter in 7 ways.
Hence, total number of ways =7×7×7×7×7=75.

Q3. In how many ways can 6 letters be posted in 4 post boxes, if any number of letters can be posted in all of the three post boxes?
a. 6C4 b. 6P4 C. 64 d. 46
correct: D, Solution:
The first letter can be posted in any of the 4 post boxes. Therefore, it has 4 choices.
Similarly, the second, the third, the fourth, the fifth and the sixth letter can each be posted in any of the 4 post boxes.
Therefore, the total number of ways the 6 letters can be posted in 4 boxes is =4×4×4×4×4×4=46

Q4. Ten coins are tossed together. What is the number of possible outcomes if third outcome was head?
A. 210 B. 29 C. 28 D. None of these
Correct: B, Solution:
22×1×27=29

Q5. In how many ways, 5 letters can be posted on 3 distinct letter boxes?
A. 35 B. 53 C. 5C3 D. 5P3
Correct: A, Solution:
First letter can be posted in 3 ways and second letter can be posted in 3 ways.
Similarly all the letters can be posted in three ways. So total no of ways =35.
‘Distinct letter boxes’ means ‘distinct boxes’, doen’t mean ‘distinct arrangements’.

Q6. How many outcomes of tosses of 10 coins are possible if third and fourth coins have resulted in heads and ninth and tenth coins have resulted in tails?
A. 32 B.64 C. 128 D. 256
correct: B, Solution:
Total number of possible outcomes 210.
Since, the outcomes of 4 of the coins are fixed.
Hence, total no of outcomes

210/24=64

Q7. How many 3-digit numbers can be made with the digits 1 to 5 if:
(i). repetitions are allowed
(ii). repetitions are not allowed
Solutions:
(i). 5 ‘boxes’ of 3 ∴ =53=125, 3-digit numbers (repetitions allowed)
(ii). 5×4×3=5!/(5-3)!=5!/2!=5P3 =60. 3-digit numbers (no repetition allowed)

Q8. How many four-digit odd numbers exist with all distinct digits?
a. 2240 b. 2520 c. 120 d. 360
correct: A, Solution:
Ones place may only be filled by each odd digit {1,3,5,7,9}. Zero (0) may not fill extreme position or thousand place here.
multiplying number of each digit filling ones place, thousands place, hundreds place and tens place in a row.
Total number of ways will be 5×8×8×7=2240

Q9. A code is made using the format XYY, where the X is any letter in the alphabet and Y represents any digit from 0 to 9.
(i). How many possible codes can be formed if the letters
and digits are repeated?
(ii). How many possible codes can be formed if the letters and digits are not repeated?
Solutions:
(i). In first slot, we have 26 possible options (26 letters in the alphabet)
In the second slot, we have 10 possible options (digits 0 to 10).
In the third slot, we have 10 possible options (digits 0 to 10 — the digits may be repeated).
∴26×10×10=2 600 possible codes
(ii). In first slot, we have 26 possible options (26 letters in the alphabet)
In the second slot, we have 10 possible options (digits 0 to 10).
In the third slot, we have 9 possible options (the digits may be repeated).
∴26×10×9=2340 possible codes

Q10. How many 3 digit numbers can be formed from the digits 1,2,3,4 and 5 assuming that
(i). repetition of the digits is allowed?
(ii). repetition of the digits is not allowed?
Solution:
(i). There will be as many ways as there are ways of filling 3 vacant palaces in succession by the given five digits. In this case, repetition of digits is allowed.
Therefore, the units place can be filled in by any of the given five digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5×5×5=125.
(ii). In this case, repetition digits is not allowed. Here, if units place is filled in first, then in can be filled by any of the given five digits.
Therefore, the number of ways of filling the units place of the three-digit number is 5.
Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given igits is 5×4×3=60

Q11. How many 3-digit even numbers can be formed the digits 1,2,3,4,5,6, if the digits can be repeated?
Solution:
There will be as many ways as there are ways of filling 3 vacant places in succession by the given six digits.
In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways.
The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.
Therefore, by multiplication principle, the required number of three digit even numbers is 3×6×6=108
alternate solutions:
Here, given 6 digits.
Three places can be filled. Therefore, repetition of digits is allowed and even numbers are to be formed. One’s place can be filled by 3 ways, ten’s place can be filled by 6 ways, hundred’s place can be filler by 6 ways.
So, by multiplication theorem of basic principle of enumeration, the number of required numbers
=3×6×6=108

Q12. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Solution:
There are as many codes as there are ways of filling 4 vacant places in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed.
The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is 10×9×8×7=5040
Hence, 5040 four-letter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.
alternate solutions:
Ten letters are given, four places can be filled, repetition of digits may not be repested.
1st place can be filled by 10 ways
2nd place can be filled by 9 ways
3rd place can be filled by 8 ways
4th place can be filled by 7 ways
The number of required ways equals

=10×9×8×7=5040

Q13: How many 5-digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?
Solution:
It is given that the 5-digit telephone numbers as there are ways of filling 3 vacant places 6.7._._._ by the digits 0-9, keeping in mind that the digits cannot be repeated.
Therefore, the units place can be filled in 8 different ways following which, the tens place can be filled in by any of the remaining 6 digits in 6 different ways.
Therefore, by multiplication principle, the require number of ways in which 5-igit telephone numbers can be constructed is 8×7×6=336

Q14: A coin is tossed 3 times and the outcomes are recored. How many people outcomes are there?
Solution 14-I:
When a coin is tossed once, the number of outcomes is 2 (head and tail) i.e., in each throw, the number of ways of showing a different face is 2.
Thus, by multiplication principle, the required number of possible outcomes is 2×2×2=8
solution 14-II:
The sample space of tossing a coin once equals 2, i.e: an image and a numeral sides. Or tail and head on another coin. So each tossing has 2 possibilities. Hence the number of required ways equals =2∙2∙2=23=8 ways.

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