1.1 Permutation
Short Questions-Answers
1.
In how many ways can
eight people be seated in a row of eight
seats so that two particular persons are always together?
Soln: Since 2 particular
persons to sit together, we consider them to be one, so there are 7 persons to
be arranged in 7 seats which can be done in 7! ways. But 2 particular persons
can interchanged their position in 2! ways.
\ Total number of arrangement = 7! ´
2!
=
7 ´
6 ´
5 ´
4 ´
3 ´
2 ´
1 ´
2 ´
1 = 1080 Ans.
2.
In how many ways can the letters of the word COMPLETE be
arranged so that the repeated letters are always together?
Soln: The
word COMPLETE contains 8 letters among which there are 2E's. Let 2E's come
together. Then, considering 2E's as a single letter, there will be 7 letters
which are C, O, M, P, L, T, (EE).
So, the total number of different
arrangements in which 2E's always come together = 7!
=
5040. Ans.
3.
In how many ways can the
letters of the word HEXAGON be arranged so that the vowels are always together
?
Soln: In
order to find the number of arrangements so that three vowels E, A, O are
always together, we may consider 3 vowels as a single letter. So, the 5
different letters H, X, G, N, (E, A, O) can be arranged in 5! ways. But the
three vowels E, A, O, can be arranged themselves in 3! ways.
\ Total
number of arrangements in which the vowels always come together
= 5! × 3! = 5 × 4 × 3 × 2 × 1 × 3 × 2
× 1 = 720 Ans.
4.
How many different numbers of five digits can be formed with
the digits 0, 1, 2, 3, 4?
Soln: The
number of five -digits numbers formed from the digits 0, 1, 2, 3, 4
= P
(5, 5) = 5! = 1 × 2 × 3 × 4 × 5 = 120.
Some of these
numbers have 0 in ten thousand's place.
If we keep 0 in ten thousand's place, one digit 0 is fixed
so there will be only 4 digits left.
\ The
number of five digits number with 0 in the first place (ten thousand' place)
P (4, 4) = 4! = 1 × 2 × 3
× 4 = 24
\ The required number
of five- digits significant numbers
= 120 – 24 = 96 Ans.
5.
How many numbers
between 3000 and 4000 can be formed with the digits 2, 3, 4, 5, 6, 7?
Soln: The required numbers must be of four digits and between 3000 and 4000. So, the digits in thousand's place must be 3 only. Thus, there is only one way of filling up the thousand's place. After filling the thousand's place, 5 digits are left. So, hundred's place can be filled up in 5 ways. After filling the thousand's and hundred's places, 4 digits are left. So, ten's place can be filled in 4 ways and the unit place by 3 ways.
\ The total numbers
that can be formed between 3000 and 4000
= 1 × 5 × 4 × 3 = 60 Ans.
6.
In how many ways can
7 students be seated in a circle?
Soln: Here, n = 7.
The possible no. of ways = (n – 1)! = (7 – 1)! = 6! = 720 Ans.
7.
In how many ways the
letters of the word ELEMENT can be arranged so that vowels are always together?
Soln: The given word
consists of three vowels namely E, E, E. considering these three vowels as one
letter, we have to arrange 5 different letters L, M, N, T, (EEE).
Here, n = 5, and the number of
arrangements in which vowels are always together = n! = 5! = 1 × 2 × 3 × 4 × 5
= 120 Ans.
0 Comments