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we can take 2 number from 5 numbers in 5c2 ways, each pair has again 2 possibilities, so 2*5c2 = 2*10 = 20 ways.
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A palindrome is a number that reads the same forward and backward. For example - 1331. How many 4-digit palindromes can be formed using two of the five digits - 1, 3, 5, 7, and 9?

A. 10
B. 20
C. 30
D. 40
E. 60

FIve digits given: 1, 3, 5, 7, and 9

2 out of 5 digits can be chosen in 5C2 = 10 ways

Palindrome for any two chosen digits a and b can be made in two ways i.e. abba and baab

Total favourable outcomes = 5C2*2 = 10*2 = 20 ways

Answer: Option B


Why are we not doing (5C2*2!)/(2!*2!) ? And when are we supposed to do (5C2*2!)/(2!*2!)(like what hint in the question suggests that we should do this ) ? Tell in case of digits or numbers also .
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Bunuel
A palindrome is a number that reads the same forward and backward. For example - 1331. How many 4-digit palindromes can be formed using two of the five digits - 1, 3, 5, 7, and 9?

A. 10
B. 20
C. 30
D. 40
E. 60

FIve digits given: 1, 3, 5, 7, and 9

2 out of 5 digits can be chosen in 5C2 = 10 ways

Palindrome for any two chosen digits a and b can be made in two ways i.e. abba and baab

Total favourable outcomes = 5C2*2 = 10*2 = 20 ways

Answer: Option B


Why are we not doing (5C2*2!)/(2!*2!) ? And when are we supposed to do (5C2*2!)/(2!*2!)(like what hint in the question suggests that we should do this ) ? Tell in case of digits or numbers also .
Bunuel or GMATinsight chetan2u


If you are doing arrangements and there are repetition of a number.

In how many ways can you write numbers from 1, 3, 3, 1 ….. four numbers can be arranged in 4! ways but two are repeated so 4!/(2!2!)

But here we are initially finding combination by 5C2 and then arranging these 2 selected numbers in 2! ways. That is why 5C2*2!
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