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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?

योग

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Combinatorics and Mathematical Induction - Exercise 4.3 [पृष्ठ १८७]

APPEARS IN

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

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