
How many different permutations can be formed from the letter of the word EXAMINATION taken four at a time?
Answer
573.3k+ views
Hint: We have to choose $4$ letters from $11$ given letter. We can do that in the following way :
(i) All four letters are different.
(ii) Two of them are alike and two are different.
(iii) Two alike of one kind and two of another kind.
Complete Step-by-step Solution
Step 1: Find $4$ letter from $11$ and all $4$ are different.
We have $8$ different types of letters i.e. A, E, I, M, N, O, T, X.
Out of these $8$ letters $4$ can be arranged in :
\[{}^8{p_4} = \dfrac{{8!}}{{4!}} = 8 \times 7 \times 6 \times 5 = 1680\] …(i)
Step 2: Two different and two alike.
Two alike letters can be chosen from one of the $3$ pairs (A,A), (I,I) and (N,N)
So total number of ways to choose one pair$ = 3$
To choose $2$ different letters we have $7$ options so total number of ways :
${}^7{C_2} = \dfrac{{7!}}{{5!2!}} = \dfrac{{7 \times 6}}{2} = 21$
Hence, total number of groups with $2$ alike and $2$ different $ = 63$
Each of the group have $4$ letter in which $2$ are same and $2$ are different and they can be arranged in themselves in $\dfrac{{4!}}{{2!}} = 12$
Hence, total number of words is $ = 63 \times 12$
$ = 756$ …..(ii)
Step 3: Two alike of one kind and two alike of another kind.
Out of three pair of letter, we have to choose two of them.
This can be done in ${}^3{C_2} = 3$ ways.
For example NNAA.
There will be arranged within the word also and they are arranged in:
$\dfrac{{4!}}{{2!2!}} = 6$ …..(iii)
Hence we have $6 \times 3 = 18$ words of this type.
Therefore, using the equation (i), (ii) and (iii). Total words with 4 letter are :
$1680 + 756 + 18 = 2454$
$\therefore $ 2454 different permutations can be formed from the letter of the word EXAMINATION taken four at a time.
Note:
In this type of question, we have to think about all the possible cases and corner cases. If any of the possible cases are missed then we might end up with the wrong answer. We also have to note that we can use permutation and combination interchangeably with some extra steps.
(i) All four letters are different.
(ii) Two of them are alike and two are different.
(iii) Two alike of one kind and two of another kind.
Complete Step-by-step Solution
Step 1: Find $4$ letter from $11$ and all $4$ are different.
We have $8$ different types of letters i.e. A, E, I, M, N, O, T, X.
Out of these $8$ letters $4$ can be arranged in :
\[{}^8{p_4} = \dfrac{{8!}}{{4!}} = 8 \times 7 \times 6 \times 5 = 1680\] …(i)
Step 2: Two different and two alike.
Two alike letters can be chosen from one of the $3$ pairs (A,A), (I,I) and (N,N)
So total number of ways to choose one pair$ = 3$
To choose $2$ different letters we have $7$ options so total number of ways :
${}^7{C_2} = \dfrac{{7!}}{{5!2!}} = \dfrac{{7 \times 6}}{2} = 21$
Hence, total number of groups with $2$ alike and $2$ different $ = 63$
Each of the group have $4$ letter in which $2$ are same and $2$ are different and they can be arranged in themselves in $\dfrac{{4!}}{{2!}} = 12$
Hence, total number of words is $ = 63 \times 12$
$ = 756$ …..(ii)
Step 3: Two alike of one kind and two alike of another kind.
Out of three pair of letter, we have to choose two of them.
This can be done in ${}^3{C_2} = 3$ ways.
For example NNAA.
There will be arranged within the word also and they are arranged in:
$\dfrac{{4!}}{{2!2!}} = 6$ …..(iii)
Hence we have $6 \times 3 = 18$ words of this type.
Therefore, using the equation (i), (ii) and (iii). Total words with 4 letter are :
$1680 + 756 + 18 = 2454$
$\therefore $ 2454 different permutations can be formed from the letter of the word EXAMINATION taken four at a time.
Note:
In this type of question, we have to think about all the possible cases and corner cases. If any of the possible cases are missed then we might end up with the wrong answer. We also have to note that we can use permutation and combination interchangeably with some extra steps.
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