Important Questions: Permutations & Combinations | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: A coin is tossed 6 times, and the outcomes are noted. How many possible outcomes can be there?
Ans: When we toss a coin once, the number of outcomes we get is 2 (Either Head or tail)
So, in each throw, the no. of ways to get a different face will be 2.
Therefore, by the multiplication principle, the required no. of possible outcomes is
2 x 2 x 2 x 2 × 2 × 2 = 64

Q2: From a team of 6 students, in how many ways can we choose a captain and vice-captain assuming one person can not hold more than one position?
Ans: From a team of 6 students, two students are to be chosen in such a way that one student will hold only one position.
Here, the no. of ways of choosing a captain and vice-captain is the permutation of 6 different things taken 2 at a time.
So, ⁶P₂ = 6! / ( 6 -2 )! = 6! / 4! = 30

Q3: How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?
Ans: No. of Vowels in the word – DAUGHTER is 3.
No. of Consonants in the word Daughter is 5.
No of ways to select a vowel = 3c2 = 3!/2!(3 – 2)! = 3
No. of ways to select a consonant = 5c3 = 5!/3!(5 – 3)! = 10
Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30
Total number of words = 30 x 5! = 3600 ways.

Q4: Find the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Ans: Take a deck of 52 cards,
To get exactly one king, 5-card combinations have to be made. It should be made in such a way that in each selection of 5 cards, or in a deck of 52 cards, there will be 4 kings.
To select 1 king out of 4 kings = ⁴c₁
To select 4 cards out of the remaining 48 cards = ⁴⁸c₄
To get the needed number of 5 card combination = ⁴c₁ x ⁴⁸c₄
= 4x² x 47 x 46 × 45
= 778320 ways.

Q5: Evaluate: 10! – 6!
Ans: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x1 = 3628800
6! = 6 X 5 x 4 x 3 x 2 x 1 = 720
10! – 6! = 3628800 – 720 = 3628080

Q6: Find the 3-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5 assuming that
(a) digits can be repeated.
Ans: 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

(b) digits are not allowed to be repeated.
Ans: By the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Q7: Evaluate the following
(i) 6 !
Ans: 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720

(ii) 5 ! – 2 !
Ans: 5! = 1 × 2 × 3 × 4 x 5 = 120
As 2! = 1 × 2 = 2
Therefore, 5 ! – 2 ! = 120-2 = 118.

Q8: How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Ans: Number of letters in word EQUATION` = 8
n = 8
If all letters of the word used at a time
r = 8
Different numbers formed = nPr
= 8P8
= 8!/(8 8)!
= 8!/0!
= 8!/1
= 8!
= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320

Q9: It is needed to seat 5 boys and 4 girls in a row so that the girl gets the even places. How many are such arrangements possible?
Ans: 5 boys and 4 girls are to be seated in a row so that the girl gets the even places.
The 5 boys can be seated in 5! Ways.
For each of the arrangements, 4 girls can be seated only at the places which are cross marked to make girls occupy the even places).
B x B x B x B x B
So, the girls can be seated in 4! Ways.
Hence, the possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Q10: Find the number of 6 digit numbers that can be formed by using the digits 0, 1, 3, 5, 7, and 9. These digits shall be divisible by 10, and no digit shall be repeated?
Ans: The number which has a 0 in its units place is divisible by 10.
If we put 0 in the unit place, _ _ _ _ 0, there will be as many ways to fill 5 vacant places. (1, 3, 5, 7, 9)
The five vacant places can be filled in 5! ways = 120.