In how many ways can 3 prizes be distributed among 4 girls, when
Question: In how many ways can 3 prizes be distributed among 4 girls, when (i) no girl gets more than one prize? (ii) a girl may get any number of prizes? (iii) no girl gets all the prizes? Solution: (i)To distribute 3 prizes among 4 girls where no girl gets more than one prizethe possible number of permutation possible are: ${ }^{4} \mathrm{P}_{3}=24$ (ii) To distribute 3 prizes among 4 girls where a girl may get any number of prizes the number of possibilities are: $4 \times 4 \times 4=64$. (S...
Read More →A customer forgets a four-digit code for an automated teller machine
Question: A customer forgets a four-digit code for an automated teller machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6, 9. Find the largest possible number of trials necessary to obtain the correct code. Solution: Given: code consists of digits 3, 5, 6, 9. To find: the largest possible number of trials necessary to obtain the correct code. The customer remembers that this 4 digit code consists of digits $3,5,6,9$. So the largest possible number of trials ...
Read More →A number lock on a suitcase has three wheels each labeled
Question: A number lock on a suitcase has three wheels each labeled with ten digits 0 to 9. if opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock. Solution: The number of sequences possible $=10 \times 9 \times 8=720($ since no repeated digits is the given condition.) There will be only one successful attempt so the number of unsuccessful attempts to open the lock ...
Read More →In how many ways can three jobs, I, II and III be assigned to three persons
Question: In how many ways can three jobs, I, II and III be assigned to three persons A, B and C if one person is assigned only one job and all are capable of doing each job? Solution: Given: three jobs, I, II and III to be assigned to three persons A, B and C. To find: In how many ways this can be done. Condition: one person is assigned only one job and all are capable of doing each job. It is given that one person is assigned only one job and all are capable of doing each job. So if for person...
Read More →How many 6-digit telephone numbers can be constructed using the digits
Question: How many 6-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: To find: 6-digit telephone numbers that can be constructed using the digits 0 to 9. Condition: each number starts with 67 and no digit appears more than once There are 10 digits between 0 to 9 , and two of them are utilized in filling up the first two digits i.e. 67 of the 6 digit phone number, so remaining number of digits $=10-2=8...
Read More →How many natural numbers less than 1000 can be formed from the digits
Question: How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times? Solution: To find: number of natural numbers less than 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times For forming a 3 digit number less than 1000 possible ways are: $5 \times 6 \times 6 \ldots$ (in 100 's place 5 digits are only possible 0 not included.) $=180$ For forming a 2 digit number les...
Read More →How many 6-digit numbers can be formed from the digits
Question: How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, 9 when no digit is repeated? How many of them are divisible by 10? Solution: There are total 6 digits available ,for forming a 6 digit number, in 100000 's place only $1,3,5,7,9$ can be used( 0 not included because it will lead to formation of 2 digit number.) In 10000 's place any of the remaining 5 digits can be used(even 0 can be used.) In 1000's place any of the remaining 4 digits can be used. In 100 's place any...
Read More →How many 3-digit numbers can be formed by using the digits
Question: How many 3-digit numbers can be formed by using the digits 0, 1, 3, 5, 7 while each digit may be repeated any number of times? Solution: 100s place 10s place Units place There are total 5 digits available, for forming a 3 digit number, in 100's place only $1,3,5,7$ can be used( 0 not included because it will lead to formation of 2 digit number.) In 10 's place any of the 5 can be used and same is the case with one's place. So total number of 3 digit numbers formed $=4 \times 5 \times 5...
Read More →How many 3-digit numbers are there with no digit repeated?
Question: How many 3-digit numbers are there with no digit repeated? Solution: In forming a 3 digit number the 100 's place can be occupied by any 9 out of 10 digits (0 not included because it will lead to formation of 2 digit number.) The 10 's place can be occupied by any of the remaining 9 digits (here 0 can or cannot be used.) In one's place any of the remain 8 digits can be used. So total 3-digit numbers with no digit repeated are: $9 \times 9 \times 8=648$....
Read More →How many numbers can be formed from the digits 1, 3, 5, 9
Question: How many numbers can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed? Solution: To find: number of numbers that can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed Forming a 4 digit number: $4 !$ Forming a 3 digit number: ${ }^{4} \mathrm{C}_{3} \times 3 !$ Forming a 2 digit number: ${ }^{4} \mathrm{C}_{2} \times 2 !$ Forming a 1 digit number: 4 So total number of ways $=4 !+\left({ }^{4} \mathrm{C}_{3} \times 3 !\right)+\left({ }...
Read More →How many 4-digit numbers are there, when a digit may
Question: How many 4-digit numbers are there, when a digit may be repeated any number of times? Solution: To find: Number of 4 digit numbers when a digit may be repeated any number of times The first place has possibilities of any of 9 digits. ( 0 not included because 0 in starting would make the number a 3 digit number.) The second place has possibilities of any of 10 digits. The third place has possibilities of any of 10 digits. The fourth place has possibilities of any of 10 digits. Since rep...
Read More →How many 3-letters words can be formed using a, b, c, d, e if
Question: How many 3-letters words can be formed using a, b, c, d, e if (i) Repetition of letters is not allowed? (ii) Repetition of letters is allowed Solution: (i) if repetition of letters is not allowed then number of many 3 -letters words that can be formed using $a, b, c, d, e$ are $5 \times 4 \times 3=60$ (ii) if repetition of letters is allowed then number of many 3-letters words that can be formed using $a, b, c, d$, e are $5 \times 5 \times 5=125$...
Read More →In how many ways can 5 letters be posted in 4 letter boxes?
Question: In how many ways can 5 letters be posted in 4 letter boxes? Solution: Each letter has 4 possible letter boxes option. So the number of ways in which 5 letters can be posted in 4 letter boxes $=4 \times 4 \times 4 \times 4$ $\times 4=4^{5}$ (Each 4 for each letter.)...
Read More →In how many ways 6 rings of different types can be worn in 4 fingers?
Question: In how many ways 6 rings of different types can be worn in 4 fingers? Solution: Given:6 rings and 4 fingers. Each ring has 4 different fingers that they can be worn. So total number of ways in which 6 rings of different types can be worn in 4 fingers $=4 x$ $4 \times 4 \times 4 \times 4 \times 4=4^{6}$...
Read More →A gentleman has 6 friends to invite.
Question: A gentleman has 6 friends to invite. In how many ways can be send invitation cards to them, if he has 3 servants to carry the cards? Solution: Given: A gentleman has 6 friends to invite. He has 3 servants to carry the cards. Each friend can be invited by 3 possible number of servants. So the number of ways of inviting 6 friends using 3 servants $=3 \times 3 \times 3 \times 3 \times 3 \times 3=3^{6}$...
Read More →Find the total number of ways of answering 5 objective-type question,
Question: Find the total number of ways of answering 5 objective-type question, each question having 4 choices. Solution: Given: 5 objective-type question, each question having 4 choices. To find: the number of ways of answering them. Each objective-type question has 4 choices. So the total number of ways of answering 5 objective-type question, each question having 4 choices $=4 \times 4 \times 4 \times 4 \times 4=4^{5}$...
Read More →In how many ways can the following prizes be given away to a class of 20
Question: In how many ways can the following prizes be given away to a class of 20 students : first and second in mathematics; first and second in chemistry; first in physics and first in English? Solution: Given: 20 students. The number of ways of giving first and second prizes in mathematics to a class of 20 students=20 19. (First prize can be given to any one of the 20 students but the second prize cannot be given to the student that received the first prize so the number of candidates for th...
Read More →For a set of five true or false questions,
Question: For a set of five true or false questions, no student has written the all correct answer and no two students have given the same sequence of answers. What is the maximum number of students in the class for this to be possible? Solution: Given: a set of five true false questions. To find: the maximum number of students in the class. Condition: no student has written the all correct answer and no two students have given the same sequence of answers. The total number of answering a set of...
Read More →A sample of 3 bulbs is tested.
Question: A sample of 3 bulbs is tested. A bulb is labeled as G if it is good and D if it is defective. Find the number of all possible outcomes. Solution: A sample of 3 bulbs is tested. A bulb is labeled as G if it is good and D if it is defective. Find the number of all possible outcomes. So number of all possible outcomes (of all bulbs) $=2 \times 2 \times 2=8$...
Read More →From among the 36 teachers in a school, one principal and one vice
Question: From among the 36 teachers in a school, one principal and one viceprincipal are to be appointed. In how many ways can this be done? Solution: Given: 36 teachers are there in a school. To find: Number of ways in which one principal and one vice-principal can be appointed. There are 36 options of appointing principal and 35 option of appointing vice-principal since same teacher cannot be appointed as principal and vice-principal. Total number of ways $=36 \times 35=1260$...
Read More →A mint prepares metallic calendars specifying months, dates and days in
Question: A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of February calendars should it prepare to serve for all the possibilities in the future years? Solution: To find: types of February calendars that can be prepared. There are two factors to develop FEBRUARY metallic calendars (1) The day on the start of the year of which possibility $=7$ (2) Whether the year is leap year or not of which possibili...
Read More →There are 6 items in column A and 6 items in column B.
Question: There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible (correct or incorrect) answers are there to this question? Solution: As we can see that For Item2 there can be any of the match So, For each item in column A there are 6 different options in column B since we don't have to think about correct or incorrect matching. So possible number of combinations possible to answer: $6 \times 5 \times 4...
Read More →How many arithmetic progressions with 10 terms are there whose first term
Question: How many arithmetic progressions with 10 terms are there whose first term in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4}? Solution: Given: Two sets: $\{1,2,3\} \\{2,3,4\}$ To find: number of A.P. with 10 n terms whose first term is in the set $\{1,2,3\}$ and whose common difference is in the set $\{2,3,4\}$ Number of arithmetic progressions with 10 terms whose first term are in the set $\{1,2,3\}$ and whose common difference is in the set $\{2,3,4\}$ are: $3 \...
Read More →Given, A = {2, 3, 5} and B = {0, 1}.
Question: Given, A = {2, 3, 5} and B = {0, 1}. Find the number of different ordered pairs in which the first entry is an element of A and the second is an element of B. Solution: This is the example of Cartesian product of two sets. The pairs in which the first entry is an element of A and the second is an element of B are : $(2,0),(2,1),(3,0),(3,1),(5,0),(5,1)$ $\Rightarrow 3 \times 2=6$...
Read More →How many 4-letter codes can be formed using the first 10 letters of the
Question: How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated? Solution: Given: first 10 letters of the English alphabet. In 4 letter code for first position there are 10 possibilities for second position there are 9 possibilities, for third position there are 8 possibilities and for fourth position there are 7 possibilities since repetition is not allowed. So total numbers of combination $=10 \times 9 \times 8 \times 7=5040$...
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