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प्रश्न
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION?
उत्तर
There are 11 fetters in the word EXAMINATION,
They are AA, II, NN, E, X, M, T, O
The four-letter strings may have
(i) 2 alike letters of one kind and 2 alike tetters of the second kind.
(ii) 2 alike fetters and 2 different letters.
(iii) All different fetters.
(i) 2 alike letter of one kind and 2 alike letters of the second kind:
There are three sets of 2 alike letters AA, II, NN.
Out of these sets, two sets can be selected in 3C2 ways.
So there are 3C2 groups each of which còntains 4 letter
strings out of which 2 are alike of one kind type and 2 are alike of the second type.
4 letters in each group can be arranged in `(4!)/(2!×2!)` ways.
Hence the total number of strings consisting of two alike letters of one kind and 2 alike letters of the second kind
= `""^3"C"_2 xx (4!)/(2! xx 2!)`
= `3 xx (4 xx 3 xx 2 xx 1)/(2 xx 2)`
= 18
(ii) 2 alike letter and 2 different letters:
Out of sets of two alike letters, one set can be chosen in 3C1 ways.
From the remaining 7 distinct letters, 2 letters can be chosen in 7C2 ways.
Thus 2 alike letters and 2 different letters can be selected in (3C1 × 7C2) ways.
There are (3C1 × 7C2) groups of 4 letters each.
Now letters of each group can be arranged among themselves in `(4!)/(2!)` ways.
Hence the total number of strings consisting of 2 alike and 2 distinct letters,
= `""^3"C"_1 xx ""^7"C"_2 xx (4!)/(2!)`
= `3 xx 7 xx 3 xx (4 xx 3 xx 2 xx 1)/(1 xx 2)`
= 3 × 21 × 12
= 756 strings
(iii) All different letters
There are 8 different letters
E, X, A, M, I, N, T, O
Out of which 4 can be selected in 8C4 ways.
So there are 8C4 groups of 4 letters each
The letter in each of 8C4 group's can be arranged in 4! ways.
∴ The total number of 4.
Letter strings in which all letters are distinct = 8C4 × 4!
= `(8 xx 7 xx 6 xx 5)/(1 xx 2 xx 3 xx 4) xx 4!`
= `(8 xx 7 xx 6 xx 5)/(4) xx 4!`
= 56 × 30
= 1680 strings
Hence the total number of 4 letter strings
= 18 + 756 + 1680
= 2454 strings
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