Permutations and Combinations : Solved Examples

Example 1: How many four-digit numbers can be formed using the digits 0, 3, 4, 5, 6, 7 if
  1. Repetition of digits is not allowed?
  2. Repetition of digits is allowed?
Sol: (i) In a four digit number, 0 cannot appear in the thousand’s place. So, the thousand’s place can be filled in 5 ways (viz. 3, 4, 5, 6, and 7). Since, the repetition of digits is not allowed and 0 can be used at hundred’s place, so the hundred’s place can be filled in 5 ways. Now, any one of the remaining four digits can be used to fill up ten’s place. So, ten’s place can be filled in 4 ways. One’s place can be filled with the remaining three digits in 3 ways. Hence, the required number of ways = 5 × 5 × 4 × 3 = 300.
(ii) For a four digit number, we have to fill up four places and 0 cannot appear in the thousand’s place. So, thousand’s place can be filled in 5 ways. Since, repetition of digits is allowed, so each of the three remaining places viz hundred’s, ten’s and one’s place can be filled in 6 ways. Hence, the required number of ways = 5 × 6 × 6 × 6 = 1,080.
Example 2: It is required to seat 5 Indians and 4 Americans in a row so that all Americans occupy the even places. How many such arrangements are possible?
Sol: In all, 9 persons are to be seated in a row and in the row of 9 positions; there are exactly four even places viz. second, fourth, sixth and eighth. It is given that these four even places are to be occupied by 4 Americans. This can be done in 4P4  ways. The remaining five positions can be filled by the 5 Indians in 5P5 ways. So, by the fundamental principle of counting, the number of seating arrangements as required is 4P4 × 5P5 = 4! × 5! = 24 × 120 = 2,880.
Example 3: Find the sum of all the numbers that can be formed with the digits 2, 3, 7, 8 taken all at a time.
Sol: The total number of numbers formed with the digits 2, 3, 7 and 8 taken all at a time = Number of arrangements of 4 digits taken = 4P4 = 4! = 24. To find the sum of these 24 numbers we will find the sum of digits at units, tens, hundred’s and thousand’s place in all these numbers. Consider the digits in the unit’s place in all these numbers. Each of the digits 2, 3, 7 and 8 occur in 3! = 6 times in the unit’s place.
So, the total ways for the digits in the unit’s place in all these numbers = (2 + 3 + 7 + 8) x 3! = 120.  Similarly, the sum of the digits in the ten’s, hundred’s and thousand’s places in all these numbers = (2 + 3 + 7 + 8) × 3! = 120 each. Hence the sum of all the numbers = (100 + 101 + 102 + 103) × 120 = 133320.
Example 4: How many words can be formed from the letters of the word ‘HALFTIME’ so that the vowels never come together?
Sol: The total number of words formed by using all the eight letters of the word ‘HALFTIME’ is 8P8 = 8! = 40,320. Now we will find the words in which the vowels are together. There are three vowels A, I and E. Let’s take them as one unit. So we have 5 letters and one unit of vowels. These 6 can be arranged in 6! ways. Also, the 3 vowels can be arranged in 3! ways. So the total number of words in which the vowels are together = 6! × 3! = 720 × 6 = 4320. So, the total number of words in which vowels are never together = Total number of words – Number of words in which vowels are together = 40,320 – 4,320 = 36,000.
Example 5: How many four digit numbers divisible by 4 can be made with the digits 1, 2, 7, 4, 9 if the repetition of digits is not allowed?
Sol:    A number is divisible by 4 if the number formed by the last two digits is divisible by 4.
So, there are four two-digit numbers divisible by 4, which can be made with the help of these digits. These are 12, 24, 72 and 92.
Now corresponding each such way the remaining three digits at thousand’s and hundred’s places can be arranged in 3P2 ways.
Hence the required number of numbers = 3P2 × 4 = 3! × 4 = 24.
Example 6: How many words can be formed using the letter X four times, the letter Y twice and the letter Z twice?
Sol: We are given 8 letters viz. X, X, X, X, Y, Y, Z, Z. Clearly, there are 8 letters of which four are of one kind, two are of second kind and two are of third kind. So, the total number of permutations is (8!/4!2!2!) =420. Hence the required number of words = 420.
Example 7: How many teams of 4 persons can be formed out of 7 men, 3 women and 5 boys if each team has a man and contains at least one woman?
Sol: The following cases can be made to form a team of four persons :
No. of Men No. of Women No. of boys No. of teams
1 1 2 7C1 × 3C1 × 5C2 = 210
1 2 1 7C1 × 3C2 × 5C1 = 105
1 3 0 7C1 × 3C3 × 5C0 = 7
Hence the number of teams = 210 + 105 + 7 = 322
Example 8: Find the number of ways in which 12 different flowers can be arranged to form a garland.
Sol: 12 different flowers can be arranged in circular form in (12 – 1)! = 11! ways. Since there is no distinction between the clockwise and anticlockwise arrangements, so, the required number of arrangements = (11!/2)
Example 9: Out of 5 boys and 2 girls, a committee of 3 is to be formed. In how many ways can it be done if at least one girl is to be included?
Sol: The committee can be constituted in the following ways:
  1. By selecting 2 boys and 1 girl.
  2. By selecting 1 boy and 2 girls
Combination Calculator: 2 boys out of 5 boys and 1 girl out of 2 girls can be chosen in 5C2 × 2C1 ways and 1 boy out of 5 boys and 2 girls out of 2 girls can be chosen in 5C1 × 2C2 ways. ∴The total number of ways of forming the committee = 5C2 × 2C1 + 5C1 × 2C2 = 20 + 5 = 25.
NMAT_Rigistration_India
Example 10:    How many different 11 letter words can be formed with the letters ppppeeeeuuk?
Sol: There are 11 letters in the given word of which 4 are p’s, 4 are e’s and 2 are u’s. The total number of words is the arrangement of 11 things, of which 4 are alike of one kind, 4 are alike of second kind and 2 are of third kind i.e.(11!/4!4!2!) . Hence, the total number of words =(11!/4!4!2!) = 34,650.