What are some good questions of class 11 permutations
Answers
Answer:
These are 6 questions Important in class 11 permutations
Step-by-step explanation:
Question 1:
Find the 2-digit numbers that can be formed from the given digits: 1, 2, 3, 4 and 5 assuming that
a) digits can be repeated.
b) digits are not allowed to be repeated.
Solution:
a) 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) 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
Question 2:
A coin is tossed 6 times, and the outcomes are noted. How many possible outcomes can be there?
Solution:
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
Question 3:
Evaluate the following
(i) 6 ! (ii) 5 ! – 2 !
Solution:
(i) 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720
(ii) 5! = 1 × 2 × 3 × 4 x 5 = 120
As 2! = 1 × 2 = 2
Therefore, 5 ! – 2 ! = 120-2 = 118.
Question 4:
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?
Solution:
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! = 2! = 2
Question 5:
How many words can be formed using all the letters of the word Dinomite, using each letter exactly one time?
Solution:
There are 7 different letters in the word – Dinomite. So the number of different words formed using these 7 letters will be 8p7 = 8! / (8 -7)! = 8
Question 6:
How many words can be formed each of 2 vowels and 3 consonants from the letters of the given word – DAUGHTER?
Solution:
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!/(3 – 2)! = 3
No. of ways to select a consonant = 5c3 = 5!/(5 – 3)! = 10
Now you know that the number of combinations of 3 consonants and 2 vowels = 10x 3 = 30
And these can be arranged in 30 x 5! = 3600 ways.