# Permutation NEB/CBSE[Class 12]

### MOST IMPORTANT QUESTIONS FOR GRADE 12 ENTRANCE EXAM                                SET-1

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Permutation
Q.1.How many ways can 4 students from a group of 15 be lined up for a photograph?
solution :
There are 15P4 possible permutations of 4 students from a group of 15.
15P4=

 15! (15−4)!

=

 15! 11!

=

 15×14×13×12×11! 11!

=15×14×13×12
=32760 different lineups

Q.2.In how many ways can a supermarket manager display 5 brands of cereals in 3 spaces on a shelf?

Solution:
This is asking for the number of permutations, since we don't want repetitions. The number of ways is:
5P3=

 5! (5−3)!

=

 5! 2!

=

 5×4×3×2! 2!

=5×4×3
=60

Q.3.How many 3 letter words can we make with the letters in the word LOVE?

Solution:
There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. The number of words is given by:

4P3=

 4! (4−3)!

=

 4! 1!

=4×3×2
=24

Q.4.Find how many ways you can rearrange the word "BANANA" all at a time?

Solution:
Given words: BANANA.
Total number of letters in "BANANA" = 6
Total number of "A" in the word "BANANA" = 3
Total number of "N" in the word "BANANA" = 2
so, the permutation =

 6! 3!×2!

=

 6×5×4×3! 3!×2!

=60

Q.5.Howny ordered arrangements are
there of the letters in the word
PHILIPPINES?
Solution:
The number of ordered arrangements of the letters in the word PHILIPPINES is:
 11! 3!1!3!1!1!1!1!

=1,108,800

Q.6.How many different ways can 3 red, 4 yellow and 2 blue bulbs be arranged in a string of Christmas tree lights with 9 sockets?
Solution:
3 red, 4 yellow and 2 blue bulbs can be arranged in a string of Christmas tree lights with 9 sockets given by:
 9! 3!×4!×2!

=1260

Q.7.In how many ways can 5 bulbs be arranged
in series?
Solution:
Since there are 5 objects, the number of ways is =5!=120

Q.8.How many different words can we make using the letters A, B, E, H and L where repetition is not allowed?
Solution:
Number of words we can make using the letters A, B, E, H and L with repetition not allowed given by: 5!=5×4×3×2×1
=120

Q.9.In how many ways can the six letters of the word "mammal" be arranged in a row?

Solution:
Since there are three "m"s, two "a"s and one "l" in the word "mammal", we have for the number of ways we can arrange the letters in the word "mammal":
 6! 3!×2!×1!

=

 6×5×4×3! 3!×2!×1!

=

 6×5×4 2!×1!

=

 6×5×4 2

=60

Q.10.If three alphabets are to be chosen from A, B, C, D and E such that repetition is not allowed then in how many ways it can be done?

Solution:
The number of ways three alphabets can be chosen from five will be,
5P3=

 5! (5−3)!

=

 5! 2!

=

 5×4×3×2×1 2×1

=60

Q.11.How many three digit numbers can be formed with the digits: 1, 2, 3, 4, 5 with repetition allowed.
Solution:
Here total number of digits n = 5.
Number of digits taken each time r = 3. We know that the number of permutations of r objects taken from n objects with repetition allowed is = nr. So number of permutations here is = 53 = 125

Q.12.Find the number of different ways of arranging 4 letters of 26-letter alphabet with repetition.
Solution:
Here total number of letters n=26. Chosen number of letters r=4.
So, The number of different ways of arranging 4 letters of 26-letter alphabet with repetition is given by: 264=456,976