Let S be the set of five-digit numbers formed by the digits 1,2,3,4 and 5,using each digit exactly once such?

Question

Let S be the set of five-digit numbers formed by the digits 1,2,3,4 and 5,using each digit exactly once such that exactly two odd positions are occupied by odd digits.What is the sum of the digits in the rightmost position of the numbers in S
(a) 228
(b) 216
(c) 294
(d) 192

Answer 1

(b) 216.

When the last digit is odd there are 16 allowable combinations. The sum of odd last digits is 5+3+1=9.

When the last digit is even there are 12 allowable combinations. The sum of even last digits is 4+2=6.

Sum of last digits = (16*9)+(12*6)
=216.

Answer 2

(b) 216

You have exactly 2 even positions. In order for exactly 2 odds digits to occupy two of the 3 odd positions you must have exactly one even position occupied by an even digit:

Note there are two permutations of the digits 4 and 2:
24 and 42.
There are 6 permutations of the digits 1,3,5
135, 153, 315, 351, 513 and 531

Case 1: The number ends in an even digit E
(O = odd)

Possibilities: OEOOE or OOOEE

There are 6 permutations of the odd digits 1,3,5. The last E can be 2 or 4 in both possibilities. So if the last digit is even then the sum of the last digit for all possibilities for each case = 6*(4+2) =36. So the total for both cases is 2 x 36 =72

Case 2: the number ends in an odd digit O

Possibilities:

EEOOO, OEEOO, OOEEO, EOOEO

Here we have 4 cases. Either E can be 2 or 4.The sum of the odd digts 1,3,5 = 9. In the 6 permutations, each digit appears twice in the last position. So the total sum of the last digits for these 4 cases will simply be 4x2x2x9 = 144

Therefore the final answer = 72 + 144 = 216

Answer 3

Wow. Let's see.

The 5-digit numbers that satisfy the question will have an odd digit in one of the even positions and an even digit in the other even position.

Thus into the "first" even position we can put any of three odd numbers, and into the "second" even position we can put any of the two even numbers, but the "first" even position and the "second" even position are permutable, so the number of 5-digit numbers with one even and one odd digit so placed is

3 * 2 * 2 = 12

Let's write them (the Xs represent digits we don't know yet):

X 1 X 2 X
X 3 X 2 X
X 5 X 2 X
X 1 X 4 X
X 3 X 4 X
X 5 X 4 X
X 2 X 1 X
X 2 X 3 X
X 2 X 5 X
X 4 X 1 X
X 4 X 3 X
X 4 X 5 X

Now, in each of those 12 5-digit numbers, the three Xs can be replaced in 6 different ways, so there are 72 5-digit numbers that satisfy the requirements of the problem. In those 72 numbers, 24 of them will have a rightmost digit of 1, 24 will have a rightmost digit of 3, and 24 will have a rightmost digit of 5, for a grand sum of 216.

The answer is (b).

Lady Venom's answer below has all three "odd" positions filled with odd digits, but the problem asks that exactly two "odd" positions be filled with odd digits. No surprise that she does not get a correct answer.

Answer 4

18 (5+1+1+3+3+5) or 196 (45+21+41+23+43+25)? how come thats not on the choices?
1 2 3 4 5
5 4 3 2 1
3 2 5 4 1
1 4 5 2 3
5 2 1 4 3
3 4 1 2 5

Answer 5

Thank you! very valuable information and this gives me better knowledge on the subject

Answer 6

hey rohit, wasting time asking question..find cat 05 solutions at some other place :)

Answer 7

Loved this question