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Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION? - Mathematics

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प्रश्न

Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION?

बेरीज
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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

shaalaa.com
Combinations
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Combinatorics and Mathematical Induction - Exercise 4.3 [पृष्ठ १८७]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 4 Combinatorics and Mathematical Induction
Exercise 4.3 | Q 21 | पृष्ठ १८७

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