Question

Question 4 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Class X1 - Maths -Permutations and Combinations Page 156

Answer 1

here, the letter of given words are A, A, E, I , I , M , N, N, O , T , X .
there are 2A, 2I, 2N, E, O , M, T , X

e.g word starting with A are formed with the letters 2I , 2N, A , E , X, M, T, O ( total 10 letters )

hence, number of words formed by these letters = 10!/2!.2!
= 10× 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2!/2!×2!
= 10 × 9 × 8 × 7 × 6 × 5 × 3 × 2 × 1
= 907200 ways .

hence, 907200 words are there in this list before the first word starting with E.

Answer 2

Answer:

907200

Step-by-step explanation:

We know that words listed before starting with E = words started with A only,

Because there is no B,C or D.

|_|_।_।_।_।_।_।_।_।_।_।

A 《 --- 10 letters--》

i.e. E,X,A,M,I-2,N-2,T,O.

Here 2 letters N & T are repeated twice.

So the no. Of words that can be formed according to given statement in the question=

10! 10×9×8×7×6×5×4×3×2

--------------- = -----------------------------------

2! × 2! 2×1×2×1

=10×9×8×7×6×5×2×3

=9,07,200

HENCE,we can create 907200 words where the words will start alphabetically before starting with E.

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