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Published on: 13/05/2022

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Questions:

Generic 2

5 x 2 = 10

How many words can be formed by using the letters of the word ORIENTAL so that A and E always occupy the odd places?

In how many ways a cricket team of eleven be chosen out of a batch of 15 players if there is no restriction on the selection?

There are six periods in each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allowed atleast one period?

In how many ways can the letters of the word PENCIL be arranged so that N is always next to E.

Write down all the permutations of the vowels A, E, I, O, U in English alphabets taking there at a time starting with A.

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Answers Key:

Generic 2

5 x 2 = 10

[Hint: There are 4 odd places in the word]
\(\therefore\) Number of permutations = ⁴P₂ = \({4!\over 2!}\) = 12
Now the remaining 6 places filled with remaining 6 letters
\(\therefore\) Number of permutations = ⁶p₆ = \({6!\over 0!}\) = 720
Hence the total number of permutations = 12 \(\times\) 720 = 8640

The total number of ways of selecting 11 players out of 15 players.
= 15C₁₁
= 15 C_15-11= 15C₄ [nC_r= nC_n-r]
\(=\frac { 15\times 14\times 13\times \times 12 }{ 4\times 3\times 2\times 1 } =1365\)

There are 6 periods in each working day of a school.
Since each subject is allowed at least one period, we first select one subject for the left out period. This can be done in 5C₁ ways.
Now, six subject can be arranged in \(=\frac { 6! }{ 2! } \) ways
Hence, total number of permutations = 5C₁ \(\times\)\(\frac { 6! }{ 2! } \)ways

= 1800.

Let us keep EN together and consider it as one letter.
Now, we have 5 letters which can be arranged in a row in 5P₅ = 5! = 120 ways.
Hence, the total number of ways in which N is always next to E is 120.

The permutations of vowels A, E, I, O, U in English alphabets taking 3 at a time and starting with A are AEI, AlE, AEO, AOE, AEU, AVE, AIO, AOI, AIU, AUI, AOU and AUO.
Clearly, there are 12 permutations.