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Hi mind helping me out for this question ?

How many way can 3 distinct letter and 2 distinct digit be arranged if the digit must be together.

Thanks.

3 Answers3

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Hi i manage to figure it out .

Let A be digit and B be letter .

Since there is 2 distinct digit and 3 distinct Letter the arrangement os

A A B B B B B B A A B A A B B B B A A B

therefor is 3!*2! * 4 = 48.

And 48 was the book answer. so is my working right ?

Thanks For ur help :)

  • Your work is correct. We can consider the block of two digits to be one object, giving us four objects to arrange. There are $4! = 24$ ways to arrange four objects (the three letters and the block of two digits) and $2!$ ways to arrange the numbers within the block, giving $4!2! = 48$ possible arrangements. It looks like you opted to arrange the three letters in $3!$ ways, arrange the two digits in $2!$ ways, and choose which of the four positions consists of the block of digits, which yields the same result. – N. F. Taussig Feb 06 '15 at 22:21
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If the two digits must be together, replace them with a fourth letter and calculate the number of permutations. Then multiply the result by a number of permutations of digits 'inside the fourth letter'.

CiaPan
  • 13,049
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Take 5 gaps. If the digits must be together, put two gaps in a box. There will be 4 ways this box can be selected. (You can check and see). Now you have 10 numbers and have to select 2 of them and arrange them. This can be done in $\binom{10}{2}2!$ ways. Then you have 26 letters and have to select 3 and then arrange them. This can be done in $\binom{26}{3}3!$. Therefore, the final answer should be $$4\binom{10}{2}2!\binom{26}{3}3!$$.